1. business and industrial
2. energy
3. electricity

BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Dear Student,
We are pleased to release the latest Exam Study Package Series in the earliest possible time. This is the
third edition of Brilliant Success’s Study Package Series(2011). We are happy to receive the great feedbacks
from our reader’s who had secured more that 90% in Physics (HSE-2010). It is our pride to get such great
achievements from our esteemed readers who had come out with flying colours in HSE-2010.
Ours is a venture in the field of academic carrers searching ways to intensify the performance of our reader’s
in Higher Secondary Examinations.
Hope you will get full benefit from our Study Package. All the best in your ensuing HSE-2011.
Publisher’s
Brilliant Success
Imphal West-1, Manipur
Send your questions to [email protected]
www.brilliantsuccess.co.cc
1
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Dear Student,
We are pleased to release the latest Exam Study Package Series in the earliest possible time. This is the
third edition of Brilliant Success’s Study Package Series(2011). We are happy to receive the great feedbacks
from our reader’s who had secured more that 90% in Physics (HSE-2010). It is our pride to get such great
achievements from our esteemed readers who had come out with flying colours in HSE-2010.
Ours is a venture in the field of academic carrers searching ways to intensify the performance of our reader’s
in Higher Secondary Examinations.
Hope you will get full benefit from our Study Package. All the best in your ensuing HSE-2011.
Publisher’s
Brilliant Success
Imphal West-1, Manipur
Send your questions to [email protected]
login to dmecino1.co.cc
2
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
HIGHER SECONDARY EXAM-2011
CRACKER
PHYSICS-XII
Unit
Marks
Unit I Electrostatics
08
Unit II Current Electricity
07
Unit III Magnetic effect of current & Magnetism
08
Unit IV Electromagnetic Induction
and Alternating current
08
Unit V Electromagnetic Waves
03
Unit VI Optics
14
Unit VII Dual Nature of Matter
04
Unit VIII Atoms and Nuclei
06
Unit IX Electronic Devices
07
Unit X Communication Systems
05
TOTAL:
70
3
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
UNIT I Electrostatics
1.1. Coulombs Law
In CGS unit, the unit of electric charge is stat coulomb(statC)
i.e. 1C  3  109 stat C
―The magnitude of force of attraction or repulsion between any
two charges is directly proportional to the product of the
charges and is inversely proportional to the square of the
distance between the charges‖
or F 
1 q1q2
where 0  absolute permittivity of free space.
4 0 r 2
(here the ch arg es are in air medium)
1 q1q2
 Fair 
4 0 r 2
and Fmed 
q1q2
1 q1q2
1
or , Fmed 
4  r 2
4 0 K r 2
The value of 0  8.854 1012 C 2 N 1m 2
Calculation gives,
1
1

 9 109 NC 2 m2
4 0 4  22  8.854 102
7
1.1. Quantisation of charge:
Quantization of ch arg e says that the amount of ch arg e possessed by a body is
q   ne ; where n  1, 2,3,.......... and e  1.6 1019 C ( smallest ch arg e that exists )
It says that value of ch arg es obtained when you put n  1, 2,3.... are accepted .
Eg. if n  2, q  2 1.6 1019  3.2 1019  valid 
ADDITIONAL SCORE
Mass of a body is affected on charging. If a body gains
electrons, its mass increases.
Properties of charges:

Charge is additive in nature.
It means that the total charge on a body can be found by
adding all the charges found at different parts of it. (Just like
adding up of 1+2+3 etc)
 The smallest amount of negative or positive charge
possessed by a body is always same.
 Charge is quantized. It means that charge on a body
is given by q   ne ; where e=charge on an
electron and n=1,2,3,......
 Unlike mass, the charge on a body is not affected by
its motion. This property is called invariance of
charge.
 The electric charge on a system is always conserved.
Eg. net charge is always same before and after a nuclear
reaction.
Charles Augustin Coulomb born in 1736 in the
Languedoc region of France
ADDITIONAL SCORE
4
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
If is the given charge, force acting on qo is
 



F  F01  F02  F03  .....  F0n =
a

i 1

F0i
1.4. CONTINUOUS CHARGE DISTRIBUTION
(1)Linear charge distribution
(2)Surface charge distribution
(3)Volume charge distribution
Linear charge density; (charge per unit length).   Q
SI unit of

l
-1
is C/m i.e Cm .
Surface charge density: (charge per surface area)   Q
A
unit of  is C/m-2 i,e C m-2.
Volume charge density :

Electric field lines
1.2. Dielectric Constant (Relative Permittivity)
K
Fair
Fmed
Re lative permttivity
r 
Fair
Fmed
r  K (dielectric cons tan t  relative permittivity )
Medium
vacuum
wax
glass
mica
water
Value of K
1
2
3 to 4
6
80
The SI unit of







Q
V
is coulomb/metre3 i.e. Cm-3
EASY SCORE
Charge on electron is - 1.6 1019 C and charge
on proton is + 1.6 1019 C .
Dielectrics are basically insulator in which
induced charges appear on their surface when an
external electric field is applied.
Coulombs law holds good for the point charge
relatively at rest.
Coulomb‘s law is applicable for microscopic as
well as very small distance of about 10-10m.
Coulombs force is independent of the mass of
the charged bodies.
Significance of Coulomb‘s law is that it gives
the nature of electrostatic force between charges.
Definition of 1 Coulomb
Here, F 
1 q1q2
4 0 r 2
You know that
1
 9  109
4 0
q1q2
r2
If q1  q2  q ; r  1m, F  9 109 N
 F  9  109
qq
r2
2
 q  1 giving q  1C
9  109  9  109
One coulomb is that charge which will repel another like
charge with a force of 9×109N when separated by a
distance of 1 metre.
1.4 . Superposition of charge
It states that ―total force acting on a given charge due to a
number of charges around it is equal to the vector sum of
forces due to individual charges‖
Numerical Problems
Q.1. How many electronic charge form one coulomb of
charge?
Soln. q = 1C and e = 1.6×10-19C
Since q = ne , giving
q
1
n 
 6.25 1018 electrons
e 1.6 1010 C
5
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Q.2. Find the force of attraction between a proton and
electron separated by a distance of 8  10
Soln. From Coulomb‘s law
14
m.
1 q1q2 .
.
4 o r 2
F
F  9  10
9

(1.61019 )2
 3.6 N
(81014 )2
(attractive)
LATEST UPDATED QUESTIONS
19
Q.1. Can a body have a charge of 0.8 10 C ?
Justify.
Ans: Charge 0.8 1019 C is smaller than the basic charge
1.6 1019 C . Hence the body cannot have charge of
0.8 1019 C .
Q.2. Two point charges
q1q2
>0 and
q1q2 <0.
and
such that
such that
What can you say about the
q1q2 >
q1q2
<0, the product is negative. It means that one
charge is +ve and the other is -ve. Hence the two charge
will attract.
Q3. The charges, initially in air medium are shifted in
water medium. Find the new forces.
Soln. Force between charges in air
In medium,
Fmed 
Fwater 
1 q1q2
Fair 
.
4 o r 2
1 q1q2
.
4  r 2
Since,   K  K 0
0
Therefore,
1.3. ELECTRIC FIELD
The region or space around a charged body within which
its influence can be felt is called electric field.
Electric field intensity
Electric field intensity: The electric field intensity at any
point in an electric field is define as the force per a unit


positive charge placed at the point. E  F . S.I unit of
q0
0 means product is +ve. It means both the
charges are negative or both are positive. Hence they repel.
If
Q. Similar charge repel each other. Can they attract
each other also?
Ans: Yes. If one is very large in magnitude as compared to
the other charge. It is due to the fact that the bigger charge
induces opposite charge such that the net charge on the
other becomes opposite.
electric field intensity is NC 1 (or Vm 1 )
electrostatic force between them?
Ans:
A soap bubble, when charged, expands when
electrified.
ADDITIONAL QUESTION
Since q1  q2  1.6 1019 C (in magnitude)
Therefore,

Hence, Fmed 
qq
1
. 1 2
4 0 K r 2
qq
1
1
1 q1q2 1
.
 Fair
. 1 2  
80 4  o r 2 80
4 0 (80) r 2
LATEST QUESTION UPDATED
Q. State limitation of coulomb‟s law.
Ans.
(1)It holds for point charges at rest.
(2)It is basically an experimental law.
(3)It loses its validity for distances less than 10-10m.
(4)It is applicable up to a few kilometer only.
(5)It is a medium dependent law.
(6)It is not a universal law.
Electric field lines
A curve or straight imaginary line followed by a unit +ve
test charge when allowed to move freely in an electric field
is known as electric field line.
Properties of electric field lines
(i) The electric field lines originate from positive charge
and terminated or end on negative charge.
(ii)The tangent at any point drawn on electric field line
represents the direction of electric field at that point.
(iii)The electric lines of force cannot intersect each other.
Reason: If two electric field lines intersect, at the point of
intersection, there will be two values of electric fields
which is not possible.
Hence they cannot intersect.
(iv)The density of electric field lines represents the
magnitude of electric field.
(v)Electric field lines are drawn perpendicular to the
surface of a positively or negatively charged sphere.
(v)Electric field lines are not allowed to pass through a
conductor as electric field inside is zero.
ADDITIONAL SCORE

Charge resides on the outer surface of a solid or
hollow conductor because similar charges repel
each other as far as possible.
6
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)


i.e. p  q(2a) or p  q(2 a ) (in vector form). Direction of
dipole moment is from negative to positive charge.
Electric field intensity at a point on the axial line of an
electric dipole
LATEST QUESTIONS UPDATED
Electric field at P due to -q at A,
1
q
EA 

 1
4   r  a 2
Q. Is it possible for electric lines of force to pass through
a dielectric or insulator?
Ans: Yes.
Electric flux
Electric flux linked with any surface is defined as the total
number of electric field lines that pass through the surface.
Flux is given by the product of magnitude of electric field
and surface area S.
 


   E. dS   EdS cos  , where   angle between E and dS
Total electric flux : The total electric flux through a
surface is defined as the surface integral of electric field.
Electric field at P due to  q at B
q
1
EB 
.
       (2)
4 0  r  a  2
As EB  E A , net electric field is given by
E  EB  E A


q
q
1
1
.

.
4 0  r  a  2 4 0  r  a  2

1
q 
 1



2
2 
4 0 
r

a
r

a







LATEST QUESTIONS UPDATED
Q. When is the total electric flux through a surface
maximum and minimum?
Ans.
   EdS cos 
 If   00 (the surface is perpendicular to the electric field )
   EdS cos(0)   EdS (max imum)
2
2 

q r  a  r  a 


2
4 0 

 r 2  a2 


 2

2
2
q  r  2ra  a  r  2ra  a 2 



2
2
2
4 0 

r  a 




q 
4ra




4 0   r 2  a 2 2 



 If   900 (the surface is parallel to the electric field )
   EdS cos(90) 0 (min imum)
1.4. Electric Dipole
A pair of two equal and opposite charges separated by a
certain distance is called an electric dipole.
 dipole length = 2a,
 Total charge on dipole q  q =0

q (2a )
2r
4 0  r 2  a 2 2

p
2r
( p  q (2a ))
4 0  r 2  a 2 2
Special case : For short dipole, a 2 can be neglected (as 2a  0)
E 
E
1 2 pr
.
.
4 0 r 4
1 2p
.
.
4 0 r 3
Electric dipole moment : It is the product of the
magnitude of either charge and dipole length.
7
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Electric field intensity at a point on the
equatorial line of an electric dipole :
1.1. Expression for potential energy stored in an electric dipole
kept in an uniform electric field
We have τ = pEsinθ
work done in rotating the dipole through a small angle dθ is
given by dW = τ dθ  net force = 0, only torque exists
To rotate from θ1 to θ 2 , the total work done is obtained by
integrating the above equation.
θ2
θ2
θ2
θ1
θ1
θ1
θ
 pEsinθ dθ = pE  sinθ dθ = pE -cos   θ2
 W =  τ dθ =
1
 pE  - cos 2 + cos 1  = -pE  cos 2  cos 1 
If θ1 = 90 then θ 2 = θ  any angle 
0
Let
 work done W = - pEcosθ. This work done is stored in the form of P.E.


E a =electric field at P due to –q at A.
E
=electric field at P due to +q at B.

/ EA / 

If θ = 0
q
1
.
       (1)
4 0  r 2  a 2 

q
1
.
     (2)
4 0  r 2  a 2 


2a
2
then P.E (U) = - pEcos 0 = - pE 1 = -pE  minimum 
0

When U is maximum, the electric dipole is in unstable
equilibrium.

When U is minimum, the electric dipole is in stable
equilibrium.

/ E / / E A / / EB /


.
BA
PA
BP

BA 
/ E / 
/ EA /
PA
r
0
q
1
.
4 0 BP 2
Since

Special Case: If θ = 1800 then P.E (U) = - pEcos 1800 = - pE -1 = pE  maximum 
q
1
.
4 0 AP 2

Also / EB / 

 P.E (U) = - pEcosθ.
b
 a2 
1
2

Expression for the torque acting on an electric dipole in a uniform
two dimensional electric field:
 PA2  a 2  r 2 
q
1
. 2


2
4 0  r  a    PA   r 2  a 2  


p
3
4 0  r 2  a 2  2
 p  q  2a 
for short dipole, a 2 is neglected .

1
P
.
4 0 r 2


Force on charge +q = + qE along the direction E .


Force on charge -q = - qE opposite to the direction of E .
These two unequal opposite forces costitutes a couple
which rotates the dipole.
 Torque = Force × arm of the couple
τ =  qE  × BC
8
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Since
sinθ =
BC
BC
=

AB
2a
 τ=
BC = 2a sinθ
 qE  × 2a sinθ
= p E sinθ
 p = q 2a  
Which is the expression for torque.


Net force is given by F
=  + qE  +  - qE  = 0 (zero net force)
net
0
Torque is maximum when θ = 90 and minimum when θ = 0
τ = p E sin 90 = p E  max 
0
and
0
τ = p E sin 0
0
= 0
 min 
1.5. Polar and Non polar molecule
Polar molecules: In polar molecules, the centre of gravity of positive charges
and the centre of gravity of negative charges does not coincide. There is no
dipole length and hence no electric dipole moment. Eg. water molecule ( A
water molecule behaves as an electric dipole).
Substances containing polar molecules are known as polar substances.
Non polar molecules: In non polar molecules, the centre of gravity of positive
charges and centre of gravity of negative charges coincide.

EASY SCORE
Since there is no separation between centre of gravity of positive and
negative charges, polar molecules have electric dipole moment .
p  q  2a 
as a  0

Since there is no separation between centre of gravity of positive and
negative charges, non polar molecules have zero electric dipole
moment.
p  q  2a   0

as a  0
Under electric field, the centre of gravity of positive and negative
charges are separated and the non polar molecules become polar
molecule.
1.6. ELECTRIC FLUX
Physical meaning:
The total number of electric field lines passing through a surface is
called electric flux.
Definition: The electric flux through a closed surface in an electric
field is defined as the surface
  integral of electric field.
Electric flux E   E . ds

Johann Carl Friedrich Gauss ( Latin:
Carolus Fridericus Gauss) (30 April 1777
– 23 February 1855) was a German
mathematician
and
scientist
who
contributed significantly to many fields,
including number theory, statistics,
analysis, differential geometry, geodesy,
geophysics, electrostatics, astronomy and
optics.
Sometimes referred to as the Princeps
mathematicorum (Latin, "the Prince of
Mathematicians" or "the foremost of
mathematicians")
and
"greatest
mathematician since antiquity," Gauss had
a remarkable influence in many fields of
mathematics and science and is ranked as
one
of
history's
most
influential
mathematicians.
He
referred
to
mathematics as "the queen of sciences."
1.7. GAUSS LAW
Gauss‘s law states that the total electric flux over a closed surface is equal to
1
0
times themagnitude of charge enclosed.
Mathematically,   q
0
9
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Proof of Gauss‟s law:
1.2. Electric field due to a line charge
(A straight charged conductor)
 
   E . ds   E ds cos 0
(E and dS are parallel) (θ=00)
 E  ds
 E  4 r 2
  dS  4 r 2 ; surface area of the sphere 
 

The electric field at the point P is given by (Refer figure of 1.12)
E
 
1
q

4  r 2
1
q
q

 4 r 2 
(Which prove Gauss ' s law)
4  r 2

1.8. Gaussian Surface: A hypothetical closed surface chosen to calculate
the surface integral of electric field is called Gaussian surface.
 Gaussian surface need not be a real physical surface.
 For calculation, Gaussian surface is considered as imaginary
surface.
1.9. To prove Coulomb‟s law from Gauss‟s law:
Here r = radius of Gaussian cylinder.
l = length of cylinder
 
   E . ds (definition

  E ds cos 0


 E ds
 E
  E  2 rl

Consider a small elementary area dS on the spherical Gaussian surface.


The electric field vector E and dS are parallel to each other and hence
θ = 0o .
Now, E =
4πo r 2
 
 =  E. dS
  =  E. dS cos θ
=  E dS = E  dS  cos 00 = 1


2
 dS = 4πr
and
cylinder.
From Gauss‘s law 
Also from Gauss's law
q
2
= E × 4πr
o

q


 2
From equations (1) and (2) we get,
q
E  2 rl 

q
 E 
2 rl 
But q   ( linear charge density)
  = E × 4πr 2  (1)
q
=
o
 (1)
 ds  surface area of the Gauss‘s
q
From definition of electric flux
Hence ,
Where
 ds
l
 E
 (2)

E=

2 r 
q
4πo r 2
If another charge q1 is kept at small portion dS, force on q 2 is given by
F = q1E
Then, F = q1
q
4πo r 2
 F=
1 qq 1
4πo r 2
which is Coulomb's law in
electrostatics.
10
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
1.10. Electric field intensity due to a uniformly
charged spherical shell:
We have,
 
 =  E . dS

If the point 
P lies
outside the surface the surface

   E . ds
 E ds cos 0
   0 
 E  ds  E  4 r 2  (1)



 ds  4 r
2



=  E dS cos θ


= E  dS ( E & dS are parallel θ = 0o )
  = E (2A)
( 
 dS = A ; 2A is for the two ends)
From Gauss's law
q
= E (2A)
o
q
σ
 E=
=
2Ao
2o
=
q
o

Also from Gauss‘s law,   q


 2
q


 σ = A surface charge density 
Special case: If the ch arg e sheet has finite thickness then,
from equation (1) and (2), E  4 r 2  q

 E
q
4  r 2
E 
2


2o
o
1.3.Electric potential at a point:
The electric potential at a point is defined as the work done in
bringing a unit positive charge from infinity up to that point.
Thus electric field intensity at any point outside a
uniformly charged spherical shell is the same as if all the
charge were concentrated at the center of the sphere.
 If the point P line on the surface of sphere, r = R
q
 E
4  R 2

If the point P lies inside the sphere, q = 0.
E=0
EASY SCORE
The electric field intensity inside a charged spherical shell
is zero.
1.11.
Electric field intensity due to infinity
plane sheet of charge:
The Gaussian surface is a cylinder with end caps each of
area A.
Total flux = flux through left end + flux through right
end.
N.B: flux through surface of cylinder is zero as electric
field is perpendicular to the area of end caps.
The electric field intensity at point A is given by
1 q
E
4 0 x 2
Small work done in moving the test charge from A to B is
dW   E dx ( work is done against the electric field)
The total work done to bring the test charge from infinity
upto P is given by
r
r
q
1
W    E dx   
2 dx
4


x
o


q r 1
q
dx  

4 o  x 2
4 o
q  1 1
q
 
 

4 o  r  
4 o r
 
r
 1
  x 

q
 Electric potential at P is given by V 
P
4 o r
11
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
POWER QUESTION
Q. Is it possible to transfer all the charges from a conductor to another insulated conductor?
Ans. Yes, it is possible. If a charged conductor is placed inside the insulated conductor so that it touches the inside wall of the
spherical conductor then the charge gets immediately transferred to the spherical conductor.
Q. Why do charges reside on the outer surface of conductors?
Ans: Due to electrostatic force of repulsion, they tend to separate from each other as far as possible and hence they stay on the
outer surfaces.
Q. What is electrostatic shielding?
Ans: It is the process of shielding an object from electric field. Eg. an object inside a hollow conductor is shielded from electric
field as electric field inside the conductor = 0.
> Behaviour of a conductor in an electric field:
(i) Inside the conductor both electric field and the charge is zero.
(ii) Charges reside only on the outer surface.
(iii) Immediately outside the surface of the conductor, the electric field is perpendicular to the surface.
(iv) The entire body of the conductor is at constant potential.
EQUIPOTENTIAL SURFACES
Any surface which has same electric potential at every point is known as equipotential surface.
Properties:
(i) No two equipotential surface can intersect each other.
Reason: If they intersect, at the point of intersection, there will be two values of electric potential which is not possible. Hence
they cannot intersect.
(ii) No work is done in moving a charge over an equipotential surface.
The work done in moving a ch arg e from
low potential VA to high potential VB is given by
WAB  q VB  VA 
For equipotential surface, VB  VA
WAB  q VB  VB   0
Hence, no work is done in moving a charge over an equipotential surface.
(iii) Equipotential surfaces help to distinguish regions of higher
field from those of weak field.
12
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Since,
dr 
1
dV  it means that the electric field
E  
E 
dr 
will be strong in the region where there is less separation
between the equipotential surfaces.
Potential Difference between two points:
Potential difference between two points is defined as the
work done per unit positive charge in bringing the charge
from lower potential to higher potential against the electric
field.
If the charge q is moved from A to B, then
Hence, the capacitance C of a capacitor can be defined as
the ratio of charge on the plates to the electric potential.
If V=1, then C= q
Therefore, the capacitance of a capacitor is numerically
equal to the electric charge require to raise its electric
potential through one unit.
SI unit of capacitance is farad.
1 farad(F)=1coulomb(C)/1 volt(V)
=1CV-1
Smaller units of capacitance:
(i) micro-farad (  F)=10-6F
(ii) milli-farad(mF)=10-3F
(iii)pico-farad (pF)=10-12F
VB  VA 
WAB
q
2.2. Parallel plate capacitor with air as medium
between the plates
1.12. Electric field as negative gradient of
electric potential:
Let a unit positive test charge +q0 be moved from the
point A to the point B against the electric field of +q.
Small work done in moving the unit test charge +q0 from
right to left
is

dW  F .dr  Fdr cos 


Since F and dr are opposite,   1800
 dW  Fdr cos(1800 )   Fdr
   (1)
q
A
The electric field between the plates is given by
If σ is thesurface charge density,  =
E

0
   (1)
 cos180  1
Also
F  q0 E and dW  q0 dV  (2)
Also E 

W
V  q 


 E
0
Where dV=potential difference between two points A and
B.
From equations (1) and (2), we get
 Fdr  q0 dV
 q0 E dr  q0 dV
E
dV
dr
Which shows that the electric field is equal to the negative
gradient of electric potential.
2.1. CAPACITORS
A device used to store maximum amount of charges is
called a capacitor or condenser.
Electric capacitance of capacitor:
It is the ability of the capacitor to store electric charge.
Definition of capacitance:
For a capacitor,
q  V  q  CV
C 
V
d
dV
(in magnitude)
dr
 as dis tan ce between the
 V  Ed 
plates  d 

d
0
Since  =
q
A
Then V 
q
q
d . Since the capaci tan ce C 
0 A
V
C
A 0
q

q
d
d
0 A
A 0
d
If a dielectric medium of dielectric cons tan t K is int roduced
Hence, C 
A 0 K
d
(completely filled by the dielectric)
between the plates, then C 
q
V
13
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
The capacitance can be increased by
(i) increasing the area A of the plates (ii) introducing a
dielectric medium of high value of K (iii) decreasing the
distance between the plates.
Dielectric strength is the maximum value of electric field
that can be applied to the dielectric without its electric
breakdown.
The dielectric constant for vacuum is 1 and its dielectric
strength is infinite.
2.5. Grouping of capacitors :
(i) capacitors in series and (ii) capacitors in parallel.
Capacitors in series:
When capacitors are connected in series, the potential
difference across each of the capacitors is different.
2.3. Capacitance of parallel plate capacitor with a
dielectric conducting medium between the plates
Therefore,
V  V1  V2  V3    (1)
q
V
q
V 
C
But C 
Here, C  A 0
0
d
E0 = original electric field ( exists over distance (d-t) )
[ V= E  d ]

[(d-t)= empty region]
=
Now
V=
C=
(d-t) [  E 0 =

0
]
1
1 1 1
  
Cs C1 C2 C3
Capacitors in parallel
q
q
 d-t   σ = 
A o
A

q
=
V
Co
 t
1

 d
C=

0
q
q
q
; V2  ; V3 
C1
C2
C3
(ch arg e on each capacitor is the same as they are in
series connection.)
q q q q
 from eq.(1),
  
C C1 C2 C3
(capacitance with air medium)
Now , V=E 0 (d-t)
So, V1 
q
q
 d-t 
A o
=
A o
A o
Co
= d
=
 t  t
 t
d  1-   1- 
 1- 
 d  d
 d
( for conducting slab)
2.4. Energy stored in charging a capacitor
dW  Vdq
 Work
q
dq
C
Total work
q

 C  V 
done in giving a ch arg e q is

q
W

0
done  potential  ch arg e 
q
q
1  q2 
dq 
C
C  2  0
1 q2
1 C 2V 2
W

 q  CV 
C 2
C 2
1
 CV 2
2
 work done is stored 
1
 Energy U  CV 2 

2
 in the capacitor as PE 
Since the capacitors are in parallel, potential difference
across each capacitor is the same ie. V. The charges
accumulated on each capacitors will be different.
 total ch arg e q  q1  q2  q3    (1)
q
 q  CV
V
 q1  C1V ; q2  C2V ; q3  C3V
Since C 
Substituting in eq. (1), we get
CV  C1V  C2V  C3V
 C p  C1  C2  C3
where C p  equivalent capaci tan ce for series connection.
The energy comes from the chemical energy
stored in the battery.
EASY SCORE
14
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)


If n number of equal capacitances are connected
in series, the equivaslent capacitance Cs =nC
If they are connected in parallel, then equivalent
capacitance
1
1 1 1
    ......n
Cp C C C

1
n

Cp C
 Cp 
C
n
Ratio of max to min . capaci tan ce
Rmax nC

 n2
C
Rmin
n
n
1000
 2.5 or n  3
400
Total capacitance of the capacitors in row is given by
1 1 1 1
   3
C 1 1 1
1
 C   F
3
Total capaci tan ce of m rows is given by
C  mC   m 
C
2

6
C  1/ 3
Therefore, he should make three rows of such capacitors,
each row containing six capacitors.
POWER QUESTIONS
Q. A capacitor of 4  F is connected to 400V supply. It
is then disconnected and connected to an uncharged
capacitor of
2  F . Calculate the common potential after the
Q.A parallel plate capacitor of capacitance C is charged
to a potential difference V and then the battery is
disconnected. Now a dielectric slab of the dimensions
equal to the spacing between the plates is inserted
between the plates. What are the changes, if any, in the
capacitance, charge, potential difference, electric field
and the energy stored ?
capacitors are connected together.
Ans. Here,
C1  4 F  4 106 F ; V1  400V and C2  2  F  2 106 F
So, ch arg e on capacitor C1 is given by
q1  C1V1  4 106  400  1.6 103 C
Ch arg e on capacitor C2 , q2  0
If C be the capaci tan ce of the combination, when C1 being ch arg ed
is connected to C2 in parallel.
6
6
C   KC (increases )
S in ce battery has been disconnected , the ch arg e on capacitor will remain same.
q
q V

 (decreases )
C  KC K
V V
E
Electric field between the plates, E    d  (decreases )
d K
K
pot. difference between the plates, V  
6
 C  C1  C2  4 10  2 10  6 10 C
Total ch arg e of combination, q  q1  q2
1.6 103  0  1.6 103 C
If V is the common potential , then,
V
Ans. Let K be dielectric constant of the slab and q, E and
U be charge on the plates of the capacitor, electric field
between the plates and energy stored in the capacitor before
inserting the slab.
On inserting dielectric slab :
q 1.6 103

 266.67V
C 6 106
Q. An electrical technician requires a capacitance of 2
 F in a circuit across a potential difference of 1KV. A
large number of capacitors are available to him, each of
which can withstand a potential difference of not more
than 400V. Suggest a possible arrangement that
requires a minimum number of capacitors.
Ans. If N capacitors are connected in m rows, each row
having n capacitors, then N=mn.
Each capacitor=1  F . Required capacitance of the
combination, C=2  F .
Voltage rating of each capacitor=400V and required
voltage rating of combination=1000V
Since the capacitors are in series, potential difference gets
added.
So, n number of capacitors connected in a row will stand a
voltage equal to 400  nV
Therefore, no. of capacitors to be connected in a row is
given by
400n=1000
1
1
V 
The energy stored in capacitor , U   C V 2  KC  
2
2
K

2
1 1 2 U
 CV   (decreases )
K2
 K
Q. Why should circuits containing capacitor be handled
cautiously, even when there is no current ?
Ans. A capacitor does not discharge itself. In case, the
capacitor is connected in a circuit containing a source of
high voltage, the capacitor charges itself to a very high
potential. If some person handles such a capacitor without
discharging it first, he may get a severe shock.
Q. A man fixes outside his house one evening a two
metre high insulating slab carrying on its top a large
aluminium sheet of area 1 m . Will he get an electric
shock, if he touches the metal sheet next morning ?
Ans. The aluminium sheet and the ground form a capacitor
with insulating slab as dielectric. The discharging current in
the atmosphere will charge the capacitor steadily and raise
its voltage. Next morning, if the man touches the metal
sheet, he will receive shock to the extent depending upon
the capacitance of the capacitor formed.
Q. If a parallel capacitor of capacitance C is kept
connected to a supply voltage V to just fill the space and
15
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
then a dielectric slab is inserted between the plates then
what will be the change in the capacitance, potential
difference, the charge, electric field and the energy
stored ?
Ans. Let K be dielectric constant of the dielectric slab and
q, E and U be charge on capacitor, electric field between
plates and energy stored in the capacitor before inserting
the slab.
On inserting dielectric slab :
The capacitance of the capacitor will become, C‘=KC
(increases)
Since the capacitor is kept connected to the supply voltage,
potential difference will remain unchanged i.e. V.
The charge on capacitor will become,
q = CV = KCV = Kq (increases)
It may be pointed out that as battery remains connected to
the capacitor, it can draw more charge from the battery.
Since, potential difference between plates does not change,
electric field will also remain unchanged.
The energy stored in the capacitor will become,
1
1
U= C V 2  KCV 2  KU (increases)
2
2
2.6. VAN DE GRAFF GENERATOR
Principle of Van de Graff generator:
(i) The discharging action of pointed ends set up an
electric wind
(ii) A charge given to a hollow conductor is
transferred to the outer surface and spreads uniformly over
it.
Construction and working: Van de Graff generator
consists of large hollow metallic sphere S mounted on two
insulating columns C and C` as shown in the Fig.
A belt runs on two pulleys P 1 and P2. The spray comb C1
is held near the lower belt which is maintained at high
positive potential of E.H.T. source.
The collector comb C2 collects the charges through its
pointed ends and transfer to the metallic sphere.
As the belt goes on revolving, the accumulation of positive
charges on the sphere also goes on increasing. Thus it can
6
generate a very high potential of the order of 5 × 10 V .
The leakage of charge is minimised by enclosing the
generator completely inside an earth - connected steel
tank. The leakage is due to the high potential on the
sphere causing ionization.
BURNING QUESTIONS
Q1. Is it possible that like charges attract?
Ans. Yes. If one of the two charges is having large
magnitude of charge than the other.
Q2. a bird perches on a bare high power line and
nothing happens. A man standing on the ground
touches the same line and gets a fatal shock. Why?
Ans. For the bird, the circuit does not get completed
between the bird and the earth and nothing happens. As for
the man the circuit gets completed and he get a fatal shock.
Q3.Is it possible to use the electricity generated during
lightning for domestic purposes?
Ans. If we could store the electricity, we can use it.
However there is no device to hold the huge electricity
generated during lightning.
Q4. What is quantization of charge? Explain why a
body cannot have a charge of 1.11019 C ?
Ans. However, the charge is not in accordance with this
law and hence it is to be considered to be invalid.
Q6. What is the smallest amount of charge that can
exist on a body?
Ans. charge on an electron (= e  1.6 1019 C )
Q7. Calculate the Coulombs force between two alpha
particles (   particles ) separated by a distance
3.2 1015 m .
Ans.
Here, r  3.2  10 15 m, q1  q2  2e  2  1.6  10 19 C
(as ch arg e on alpha partice  2e)
F 
1 q1q2
(2 1.6 10 19 ) 2
. 2  9  109 
 90 N
2
4 0 r
 3.2 1015 
16
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Q8. Write down the value of absolute permittivity of
free space.
Ans. 0  8.854 1012 C 2 N 1m2
Q16. Briefly discuss the principle, construction
working of a Van de Graff Generator.
Q9. How are permittivity and dielectric constant (or
relative permittivity) related?
Ans.
Q17. What meaning would you give to the capacitance
of a single conductor?
Ans. A single capacitor also possesses capacitance. It is a
capacitor whose one plate is at infinity.
0 K or 0r (r  K  dielctric cons tan t )
Q10. What is meant by saying that dielectric constant
for water is 80?
Ans. It means that the electrostatic force between the
chages reduces to 1/80 th times when placed in water
medium.
Q11. Why one ignore the quantization of charge when
dealing with macroscopic (large charges) charges?
Ans. In practice, the charges on bodies are large whereas
the charge on electrons are smaller. If electron (of charge e)
is added or removed from a charged body, there is not
much change on the charge of the body. Hence while
dealing with large amount of charges, quantization of
charge is ignored.
Q12. Two identical point charges Q are kept at a
distance r from each other. A third point charge -q is
placed on the line joining the two charges such that all
the three charges are in equilibrium. What is the
magnitude, sign and position of the third charge?
For ch arg e  q to be in equilibrium,
force on  q due to  Q at A  froce on  q due to  Q at B
1
qQ
1
qQ
.

.
4 0 x 2
4 0 ( r  x) 2
 x 2  (r  x) 2
x 
r
2
For  Q to be in equilibrium,
force on  Q at A due to  q  force on  Q at B due to  q
1
qQ
1
qQ

.

.
4 0  r  2
4 0 r 2

2

q
Q18. Is there an electric field inside a conductor?
Ans. No. The electric field inside a conductor is zero.
Q19. Two copper spheres of same radii, one hollow and
the other solid are charged to same potential. Which, if
any, of the two sphere will have more charge?
Ans: Same.
Q21. Why is the Van de Graff Generator enclosed
inside a steel tank filled with air pressure?
Ans. To prevent leakage of charge due to ionization.
Q22. What is Gaussian surface? What is its use?
Ans. Any closed surface around the charge so that Gauss‘s
law can be applied successfully to find the electric field
intensity is known as Gaussian surface.
It is used to find surface integral of electric field .
Q23. If Coulomb‟s law involved 1 dependence, would
r3
Ans.

and
Q
4
So, the system will be in equilibrium if charge -q equal to Q is placed at the mid point of the distance between the
4
charges +Q.
Q13. Define capacitance. Derive an expression for the
capacitance of a parallel plate capacitor.
Q14. Derive an expression for the energy stored in a
capacitor.
Q15. Explain how does the capacitance of a capacitor
gets modified, when a dielectric slab is introduced
between the plates.
Gauss‟s law be valid?
Ans. No. Gauss‘s law is valid for inverse square laws only.
( 1 )
r3
Q24. What is difference between a sheet of charge and a
plane conductor having charge?
Ans. On a sheet of charge, the same charge is found on
both the surfaces while in a plane conductor, charge on
each side is different.
Q25. A man inside an insulated metallic cage does not
receive any electric shock when the cage is highly
charged, why?
Ans. Since the electric potential is the same everywhere
inside a metallic cage, no potential difference is created
inside and the man does not get any shock.
Q26. State Gauss‟s theorem. Find an expression for the
electric field due to an infinitely long line charge.
Q27.State Gauss‟s theorem in electrostatics. Apply this
theorem to calculate the electric field due to an infinite
plane sheet of charge.
Q28. Applying Gauss‟s theorem show that for a
spherical shell, the electric field inside a shell vanishes,
wheras outside it, the electric field is as if all the charge
has been concentrated at the centre.
17
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Q29. How does a dielectric differ from an insulator ?
Ans. Both the dielectrics and insulators cannot conduct
electricity. However, in case of a dielectric, when an
external field is applied ; induced charges appear on the
faces of the dielectric. In other words, dielectrics have the
property of transmitting electric effects without conducting.
Q30. Explain why presence of a dielectric increases
capacitance of the capacitor ?
Ans. When a dielectric slab is introduced between the two
plates of a capacitor, the electric field between the plates
gets reduced due to polarisation of the dielectric. The
reduced value of electric field is equivalent to a decreased
value of potential difference between the plates. In order to
make the potential difference again same, more charge has
to be given to the capacitor. i.e. the capacitance of the
capacitor increases.
Ans. If the electric field is uniform, the net force is equal to
zero. If the field is non uniform, the net force is not equal to
zero.
Q36. What is the ratio of the strength of electric field at
a point on the axial line at a point at the same distance
on the equitorial line of an electric dipole of very small
length?
Ans. As we know that
Eaxial 

1 2p
1
p
.
and Eequatorial 
.
4 0 r 3
4 0 r 3
Eaxial
Eequatorial
1 2p
.
4 0 r 3

2
1
p
. 3
4 0 r
ADDITIONAL POWER QUESTIONS
Q31. Is,there any kind of material that when inserted
between the plates of a capacitor reduces its capacitance
? f instead of a dielectric slab, we put a slab of metal
between the plates of a capacitor keeping it insulated
from them, what effect does it have on capacitance ?
Ans. The dielectric constant K of a material is always
greater than 1 and is defined as the ratio of the capacitance
(C) of the capacitor with dielectric between its plates to its
capacitance (C‘) without the dielectric between the plates,
Thus,
C

K
C
As K  1, C  C
. ie. capacitance of a capacitor on placing the dielectric
between the plates is always greater than the capacitance
without dielectric.
Q32.What limits the maximum potential to which the
hollow sphere in a Van de Graaff generator can be
raised ?
Ans. When the rate of loss of charge because of leakage
due to ionisation of surrounding air becomes equal to the
rate at which the charge is transferred to the sphere, the
maximum potential of the sphere is reached.
Q33. If instead of a dielectric slab, we put a slab of
metal between the plates of a capacitor keeping it
insulated from them, what effect does it have on
capacitance ?
Ans. When a slab of metal is put between the plates of a
capacitor, its capacitance increases, provided the slab does
not touch the plates of the capacitor. In case, the slab of
metal touches the two plates, both plates become at the
same potential and as a result, the capacitance of the
capacitor becomes zero.
Q34. Is it correct to write the unit of electric dipole
moment as mC?
Ans. Wrong. mC stands for milli coulomb.
Q35. What is the net force experienced by an electric
dipole in an uniform electric field? What about if the
electric field is non uniform?
Q1. Obtain an expression for electric field due to a
point charge.
Q2. Find the expression for the electric field intensity at
a point on the (i) axial line and (ii) equatorial line of an
electric dipole.
Q3. What is electric potential energy due to a system of
two charges? Find the expression for it.
Ans. The electrostatic potential energy of two point
charges can be defined as the work done required to bring
the charges constituting the system to their respective
positions from infinity.
1 q1q2
Mathematically,
U
.
4 0 r12
Q4. An electrostatic field line cannot have sudden
breaks. Why?
Ans. If the electrostatic field line have sudden breaks, it
will indicate absence of electric field at that point in the
electric field which is not possible. Hence, the line cannot
have sudden break.
ADDITIONAL EXAMINATION QUESTIONS-2011
Q1.Define relative permittivity of a medium.
A. Relative
permittivity (dielectric constant)

k ( r ) =
0
where  = permittivity and 0  absolute permittivity of free space
Q2. What is meant by quantization of charge and
conservation of charge ?
A. Quantization of charge: The charge on a body is an
 e.
integral
multiple
of
ie. q =  ne where e = charge on an electron = 1.6  10-19C
.
Conservation of charge The total charge on an isolated
body is always conserved ie. total number of charge
before and after a chemical reaction are the same.
Q3. What does q1 + q2 = 0 signify ?
18
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
A.
q1 + q2 = 0 means that q1 = - q2 Such a system
is an electric dipole where the two charges are
equal and opposite.
Q4. What is the nature of symmetry of the dipole field ?
A. The dipole field has a cylindrical symmetry. The axis of
the cylinder passes through the dipole axis.
Q5. When is an electric dipole in unstable equilibrium
in an electric field?


Q8. At what points, dipole field intensity is parallel to
the line joining charges?
A. At any point on axial line or equatorial line of
the dipole.
+3μC each are 100 cm
Q9. Two point charges of
apart. At what point on the line joining charges will the
electric intensity be zero?
A. The electric intensity will be zero at the centre of the
point charges.
Q10. Derive an expression for electric field intensity at a
point due to a point charge.
A. Assume that a unit positive test charge is kept at P.
Force on +q0 at point
1
4πo
qq
1
o
F=
4πo
r2
qq
r2
o
If V = constant, then E = -
.
means that when electric potential is constant, electric field
in that region is zero.
Q15. How many electrons volt make one joule ?
Ans.
Here, 1eV = 1.6 × 10
 Electric field is the force per charge 
o
E =
-19
J
 1J =
1
19
eV = 0.625×10 eV
1.6 × 10-19
Q16. An electric dipole of dipole moment 20 X 10 -6 Cm
is enclosed by a closed surface. What is the net flux
coming out of the surface ?
Ans. Since net charge on the electric dipole is zero, net flux
coming out of the surface will be zero.
Q17. A technician has only two capacitors. By using
them singly, in series or in parallel, he is able to obtain
the capacitance of 4, 5, 20, and 25  F. What are the
capacitance of the two capacitors ?
Ans. Maximum capacitance is obtained for parallel
combination. Therefore
Cp = 25  F. The
minimum capacitance is obtained for series combination
ie. Cs = 4  F. The remaining values 5  F and 20  F
will represent individual values of the two capacitors.
Q18. If two isolated conductors having definite capacity
are separated by a fine wire. Calculate the capacitance.
Are they
in series or parallel combination ?
A. On connecting, both the spheres acquire a common
potential V. The total charge will be given by
q1 + q2 = CV
( since
q = CV)
 C=
F
Now, E =
q
dV
d
= (constant) = 0  E = 0 . It
dr
dr
0
ie.  = 180
Ans. When p is antiparallel to E
Q6. Is torque on an electric dipole a vector?
Ans. Torque is a vector quantity.
Q7. Give the SI unit of electric dipole moment.
A. It is C-m ( coulomb- metre)
P is F =
Ans.
1 q qo 1
4πo r 2 q
o
=
1 q
4πo r 2
q1
V

q2
V
 C1  C2
Hence, they act as
parallel combination.
Q19. Sketch a graph to show how charge Q given to a
capacitor of capacity C varies with the potential
difference V.
A. The graph is a straight line.
Q11. If a point charge be rotated in a circle of radius r
around a charge q, what will be the work done ?
Ans. The circle of radius r will act as an equipotential
surface and hence no work is done in moving the charge.
Q12. What is the amount of work done in moving a
220C
charge between two points 5cm apart on an
equipotential surface.
A. On an equipotential surface, work done in moving the
charge is zero.
Q13. Do electrons tend to go regions of high potential
or low potential ?
A. Since electrons are negatively charged, they have
tendency to go to regions of higher potential.
Q14. If V equals a constant throughout a given region
of space, what can you say about E in that region?
Q20. Can you place a parallel plate capacitor of one
farad capacity in your house ?
A. No.
19
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
The capacitor will become too large.
Ao
d
Since C =
 A   Cd
o
A
Cd
.
o
-3
If we take d = 1mm = 10 m
 A=
1  10-3
8 2
 10 m
8.854 1012
Q29. State Coulombs law of forces between two charges
at rest. What is the force of repulsion between two
charges of 1C kept 1m apart in vacuum?
Ans.
According to Coulomb's law, force between two charges
q and q is given by
1
2
1
F α q1q 2 and F α
r2
This is a very big size to accommodate in a room!
where r = distance between the charges.
Q21. Can there be a potential difference between two
conductors of same volume carrying equal positive
charges?
A. Yes, because two conductors of same volume may have
different shapes and hence different capacitances.
Q22. A parallel capacitor has a capacity of 6  C in air
and 60  C when dielctric medium is introduced.
q q
1 q 1q 2
F α 1 2  F =
2
4πε 0 r 2
r
1
where
= constant = 9×109 Nm 2C-2
4πε 0
What is the dielectric constant of the medium ?
A.
Di-electric constant K =
If q1 = q 2 = 1C ; r = 1m
9 1×1
9
F = 9×10 ×
 F = 9×10 N
12
Cm capacitance in medium
60
=
=
 10
Co
capacitance in air
6
Q23. Where does the energy of a capacitor reside ?
A. The energy resides in the di-electric medium separating
the two plates.
Q24. Why does the electric field inside a dielectric
decrease when it is placed in an external electric field ?
A. It is because the dielectric gets polarised.
Q25. How much work must be done to charge a 24 F
capacitor, when the potential difference between the
plates is 500 V?
A.
Here, C = 24 μF = 24 × 10
 Work done =
-6
F and V = 500 Volts.
1
1
CV 2 = ( 24 × 10-6 ) 5002 = 3 joules
2
2
Q26. By what factor does the capacity of a metal sphere
increase if its volume is tripled ?
A.
1
1
1
 V 3
3
C2 R 2  V2  3
=
=
  C 2 = C1  2  = C1  3  1.44 C1
C1
R1  V1 
V
 1
Q27. What are dielectric substances ?
A. Dielectric substances are basically insulators on which
electrical effects can be passed through them without actual
conduction(charges will develop on the dielectric when
electric field is applied).
Q28. An uncharged insulated conductor A is brought
near a charged insulated conductor B. What happen to
charge and potential ?
A. Charge on B remains same, but the potential of B gets
lowered because it induces charge of opposite sign on
conductor A.(when a body induces charge on other body,
its potential will be lowered)
39. Can you place a parallel plate capacitor of one
farad capacity in your house ?
A. No.
This is a very big size to accomodate in a room!
Q 30. Two insulated charged spheres of radii 10 cm and
20 cm having same charge are connected by a
20
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
conductor and then they are separated. Which of the
two spheres will carry more charge ?
A. Bigger sphere will carry more charge as its capacity is
larger, ( q = CV). The potential V becomes same on
connecting them with a wire.
Q 31. Can there be a potential difference between two
conductors of same volume carrying equal positive
charges ?
Ans. Yes. because two conductors of same volume may
have different shapes and hence different capacitances.
Q 32. What is the net charge on a charged capacitor ?
Ans. Zero, because one plate has positive charge and the
other carries an equal negative charge.
Q 33. If the plates of a charged capacitor be suddenly
connected to each other by a wire, what will happen ?
Ans. The capacitor will be discharged immediately.
Q 34. On which factors does the capacitance of a
capacitor depend ?
Ans. It depends on geometry of the plates, distance
between them and nature of dielectric medium separating
the plates.

Electric Field

The test charge used to measure electric field at a
point has to be vanishingly small. If the test
charge is not vanishingly small, then the test
charge may bring about a change in electric field
at the observation point.
For this, one is advised to understand the difference
between the statement „the force experienced by a unit
positive test charge‟ and „the force experienced per unit
positive test charge.‟ The former statement is incorrect, as a
unit positive charge used as test charge will cause a change
in the value of electric field at the observation point.




Q 35. What is the basic use of a capacitor ?
Ans. To store charge and energy.

Q 36 . What meaning would you give to capacity of a
single conductor ?
Ans. A single conductor can be visualised as a capacitor
whose second plate is far away at infinity.


Q37. Why does the electric field inside a dielectric
decrease when it is placed in an external electric field?
Ans. It is because the dielectric gets polarised.

Q 38. Where does the energy of a capacitor reside ?
Ans. The energy resides in the dielectric medium
separating the two plates.
FACT FILE FOR
EXAMINATIONS
MAXIMUM
SCORE

IN
Coulombs Law
 The charges developed on the bodies during the
process of rubbing are due to the transfer of
electrons only from one body to the other.
 Coulomb‘s law in electrostatics holds only for
stationary charges and the charges, points in size.
 When the same two charges located in air are
placed in a dielectric medium without altering the
distance between them, electrostatic force always
decreases.
 A system of charge is said to be in equilibrium, if
the net force experienced by each charge of the
system is zero.
Electrostatic force between two charges is not
affected by the presence of a third charge in their
vicinity.
The unit of electric dipole moment is C m and not
m C. It is because, m C is used as mill coulomb.
The direction of electric field at a point on the
axial line of an electric dipole is same as that of
p(dipole moment).
The direction of electric field at a point on
equatorial line of an electric dipole is opposite to
that of p.
For very large distance of observation point from
the charged circular loop (x > > a), the circular
loop behaves as a point charge.
No torque acts on a dipoie, when it is aligned
along the direction of electric field.
The electrostatic potential energy of the dipole is
maximum, when it is aligned anti parallel to the
direction of the electric field.
The electric lines of force do not exist but what
they represent is a reality.
It gives the path along which positive test charge
would move, when free to do so.
In a uniform electric field, the lines of force are
parallel lines. The electric line of force crowd
near each other in regions of strong electric field.
The relative closeness of the lines of force at
different points in the electric field gives an
estimate of the strength of the electric field at
these points.
Electric Potential

The work done in moving a test charge between
two points in an electric field is independent of
the path followed between the two points.

The work done in moving a test charge over a
closed path in an electric field is always zero.
The above two results hold for the reason that
electrostatic force between two charges obeys
inverse law i.e. electrostatic field is inverse
square field.
21
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
> > The work done per unit positive test charge from a
point A to point B is equal to VB — VA
i.e. potential difference between the points B and A (and
not between points A and B).

Electric potential at any point inside a charged
spherical conductor is always equal to that at a
point on its surface i.e. it is same (constant) every
where.
As a result of it, the electric field inside a charged spherical
conductor is zero.

For a point outside the charged spherical
conductor, the whole charge appears to be
concentrated at its centre.

The electrostatic potential increases steadily as
one moves against the direction of electric field
and it decreases in the direction of electric field.
The above result is a consequence of negative sign in the
relation:
dV
E
dr

The electric field is always at right angle to the
equipotential surface.

No work is done in moving a test charge between
two points on an equipotential surface,

.If work has to be done in moving one of the
charges of a system from one point to some other,
then electrostatic potential energy of the system
increases by an amount equal to the work done in
moving the charge.
2.7. Gauss‟s Theorem
>> The electric flux through a small portion of the closed
surface is affected by the charges present outside the
surface.

The total electrical flux through a closed surface
is not affected by the charge present outside the
surface. It depends only upon the charges
enclosed by the close surface.
2.8. Capacitors

The capacitors are connected in parallel to
increase the capacitance.

The capacitors are connected in series to decrease
the capacitance

When a dielectric slab of dielectric constant K is
introduced between the plates of a charged
air-capacitor (battery disconnected after charging), then
(a) capacitance increases by a factor K,
(b) potential difference decreases by a factor K.
(c) charge oh the capacitor remains unaffected.
(d) electrostatic potential energy stored in the capacitor
decreases by a factor K.
> When a dielectric slab of dielectric constant K is
introduced between the plates of an air capacitor (battery
remains connected across the capacitor), then
(a) capacitance increases by a factor K
(b)
potential difference across capacitor remains
unaffected.
(c) charge on the capacitor increases by a factor K.
(d) electrostatic potential energy stored increased by a
factor K.

The dielectric slab can be removed from the
space between the plates of a charged capacitor
only by performing work. However, the work
done appears as increase in the potential energy
of the capacitor.

When a part of the space between the two plates
of a capacitor is filled with the slab of one
dielectric slab, then the two parts of the capacitor
may be looked upon as the combination of two
capacitors.

The capacitor behaves as a parallel combination
of the two sub capacitors, if each slab is of
thickness equal to separation d. On the other
hand, it behaves as a series combination of the
two sub capacitors, if the slabs together make
thickness equal to d.

The electric flux through a small portion of the
closed surface may change, if the charges inside
the closed surface are moved to the new
positions.

The total electric flux through a closed surface is
not affected, if the charges inside the closed
surface are moved to new positions.

For both the spherical shell and the sphere,
electric field at a point outside is same as if the
charge is . concentrated at their centre.
UNIT II

Inside a spherical shell, electric potential is zero.
It is zero inside a sphere also, provided sphere is
conducting.
Current electricity

A sphere of charge is not a conducting sphere.
22
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Electric current: The flow of charges in a definite direction
is called electric charge.
q
Electric current, I  . If the rate of flow of charge caries
t
with time, then current at any time (instantaneous current)
is given by i  dq ; where dq is a small charge passing
Electric current is a scalar quantity
Since both charge (q) and time (t) are scalars, electric
current is a scalar quantity. I  q
t
dt
through any cross-section of the conductor in small time.
Unit of electric current: The SI unit of current is coulomb
per second (Cs-1).
One ampere of current: One ampere of current is said to
flow through a conductor if at any cross-section, one
coulomb
of
charge
flows
in
one
second.
q
t
If q  1C and t  1sec then I  1ampere
Since I 
Types of electric current
(i)
Varying current: When the magnitude of
current does not change with time it is
called a steady current.
(ii)
Varying current: When the magnitude of
current changes with time, it is called a
varying current.
(iii)
Alternating current: If the current changes
its magnitude continuously with time and
the direction changes periodically, the
current is called alternating current.
Current Carriers
(i)
(ii)
(iii)
Current carriers in solid conductors: In solid
conductors, there are a number of free
electrons. If electric field is applied to the
conductor, the free electrons start drifting in
a direction opposite to the electric field. It
means that free electrons are the current
carriers in solid conductors.
Current
carriers
in
liquids:
In
electrolyte(conducting liquid), the positive
ions (e.g. Cu++) and negative ions (e.g. SO4-). When electric field is applied the positive
ions move in one direction and the negative
ions move in opposite direction to constitute
electric current. Therefore, in conducting
liquids(electrolytes), the ions (positive and
negative) are the current carriers.
Current carriers in gases: The electrons and
positive ions are the current carriers in
gases.
Ohms law
Ohm‘s law gives the relationship between voltage across
and current through a conductor. It was discovered by
German scientist George Simon Ohm.
It states that “ the current (I) flowing through a conductor
is directly proportional to the potential difference (V)
across its ends provided the physical conditions
(temperature, strain, etc.) do not change”.
Mathematically,
I V
or
V
 cons tan t  R
I
Where R is a constant of proportionality and is called
resistance of the conductor.
If a graph is drawn between applied potential difference
(V) and current (I) flowing through the conductor, the
graph will be a straight line passing through the origin.
23
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Factors upon which resistance depends
The resistance of a conductor depends upon the length, area
of cross section, nature of material and change in
temperature.
2.9. Ohmic and non ohmic substances
Those conductors (e.g. metals) which obey Ohm’s law are
called ohmic substances. For these conductors, the V-I
graph is linear. If Ohm’s law is not obeyed, the
substances are known as non ohmic substances. (e.g.
semiconductors).
The resistance of a conductor is
(i)
(ii)
(iii)
Resistance of a conductor
Resistance of a conductor is defined as the ratio of
potential difference applied across its ends to the resulting
(iv)
Since, R  l
1
A
l
R 
A
l
R
A
R
current through the conducto r .i.e. R  V
I
Resistance is the opposition offered by the substance to the
flow of electric current. The cause of resistance is due to
atoms and molecules of the substance. The atoms and
molecules obstruct the flow of charges (free electrons
here).
Unit of resistance
V , the S.I. unit of potential difference is volt
Here, R 
I
(V) and that of current is ampere (A). Hence the S.I. unit of
current is
directly proportional to its length
inversely proportional to the area of cross
section of the conductor
dependent on the nature of the material of
the conductor
dependent on the change in temperature.
VA1 .
Definition of one ohm(1 Ohm): A conductor is said to
have a resistance of 1 Ohm if a potential difference 1V
across its ends causes a current of 1A to flow through it.
Where
 (Greek letter ‗Rho‘)is a constant of
proportionality and is known as resistivity or specific
resistance of the conductor, Its value depends upon the
nature of the material and temperature.
2.10. Resistivity or specific Resistance:
We have seen above that R   l
A
If l  1m; A  1m 2 , then R  
Specific resistance (or resistivity) 0f a material is the resistance offered by 1 m length of wire of the material having area of
2
x-section of 1m .
Q. A length of wire has a resistance of 4.5 ohms. Find the resistance of another wire of the same material three times as
long and twice the cross-sectional area.
Solution:
For the first case,

R1  4.5  ;
l1  l;
A1  A
R2
l
A
 3l   A 
 2  1    
  1.5
R1
l1
A2
 l   2A 
R2  1.5 R1  1.5  4.5  6.75
For the second case, R2  ?;
l2  3 l;
A2  2 A
24
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Now R   l1 ;
1
A1
R2  
l2
;
A2
Q: A copper wire of diameter 1 cm has a resistance of 0.15. It is drawn under pressure so that its
diameter is reduced to50%. What is the new resistance of the wire?
Ans
 2
(1)  0.785cm 2
4

Area of wire after drawing , A2  (0.5) 2  0.196cm 2
4
Volume remains same before and after drawing
 initial volume  final volume
l
A 0.785
A1l1  A2l2 or 2  1 
4
l1 A2 0.196
Area of wire before drawing , A1 
For the first case, R1  0.15 and A1  0.785cm 2 ; l1  l
For the sec ond case, R2  ? ; A2  0.196cm 2 , l2  4l
Now, R1  
l1
l
; R2   2
A1
A2
R2  l2   A1 
       4  4  16
R1  l1   A2 
or , R2  16 R1  16  0.15  2.4

Q. A copper wire is stretched so that its length is increased by 0.1%. What is the percentage
change in its resistance?
Solution: Suppose the initial length and area of copper wire are
values be l

'
l'  1 
and A
'
l and A. Let after stretching, these
respectively.
0.1
 l  1.001l
100
Since volume of wire remains the same before
and after stretching , Al  Al 
l
A
1 
 l

 
l  1.001  l  1.001 
Let R and Rbe the resis tan ces of wire before and
after stretching ,
l
l
R   ; R  
A
A
R  l   A 

  
  1.001 1.001  1.002
R
 l    A 
R  R
Percentage increase 
 100  0.002 100
R
 0.2%
or A  A
25
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
2.11. CARBON RESISTORS
Colour code for carbon resistors: Since a carbon resistor is physically quite small, it is more convenient to use a colour
code indicating the resistance value than to imprint the numerical value on the case. In this scheme, there are generally four
colour bands A, B, C and D printed on the body of the resistor as shown in Fig. The first three colour bands (A, B and C)
given the value of the resistance while the fourth band (D) tells about the tolerance in percentage. The table below shows the
colour code for resistance values and colour code for tolerance.
Eg. For colour bands green, brown, yellow and gold, resistance =( 51×10
gold=  5%)
4
 5%) Ω (green=5, brown=1, yellow=104 and
2.12. MECHANISM OF CURRENT CONDUCTION IN METALS
When potential difference is applied across the ends of a conductor (say copper wire) as shown in Fig: electric field is
applied at ever point of the copper wire. The electric field exerts force on the free electrons which start accelerating towards the
positive terminal (i.e., opposite to the direction of the field). As the free electrons move, they collide again end again with
positive ions of the metal. Each collision destroys the extra velocity gained by the free electrons.
  . Its value i s of the

The average time that an electron spends between two collisions is called the relaxation time

order of 10 second
The average velocity with which free electrons get drifted in a metallic conductor under the influence of electric field is
14

called drift velocity
 v  .The drift velocity of electrons is of the order of 10
d
5
ms 1
26
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
2.13. RELATION BETWEEN ELECTRICFIELD AND DRIFT VELOCITY
Consider a metallic conductor connected to a battery as
shown in Fig:
2.14. RELATION BETWEEN CURRENT AND
DRIFT VELOCITY
Consider a portion of a copper wire through which current I
is flowing as shown in Fig of 2.13.
Clearly, copper wire is under the influence of electric field.
Let
l = length of the conductor
v = p.d. across the conductor
m = mass of electron
e = charge on electron

vd = drift velocity of the free electrons
Let A = area of X- section of the wire
n = electron density, i.e., number of free
electrons per unit volume
e = charge on each electron

vd = drift velocity if free electrons
 = relaxation time
Magnitude of electric field, E  V l
Under the influence of electric field, each free electron
experiences a force of  eE . The acceleration a of the
So, volume  Al  Avd
electron is given by:

eE
a
m
Since relaxation time is
length(l )
time(t )
If t  1s, vd  l
vd 
 , the drift velocity of the free
electron is given by;

 
eE 
vd  a   
m


eE 
 vd  
m
The negative sign shown that direction of drift velocity is
opposite to that of the electric field. Note that velocity is
directly proportional to the applied electric field.
Total number of electrons in volume Avd  ( Avd )n
where n  electron density
Therefore, total charge=  nAvd  e . Therefore, a charge of
neAvd per second passes the cross-section at P.

I  n e Avd
since A, n and e are constant, I  vd
Hence, current flowing though a conductor is directly
proportional to the drift velocity.
Note: The drift velocity of free electrons is very small.
Since the number of free electrons in a metallic conductor
is very large, even small drift velocity of free electrons
gives rise to sufficient current.
27
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
2.15. VARIATION OF RESISTIVITY WITH
TEMPERATURE
2.16. EFFECT OF TEMPERATURE ON RESISTANCE
Consider ametallic conductor having resistance R 0 at 0C
R
and R1 at t1 C. Then in the normal rang of temperatures,
V
ml
m l 


 
I
n Ae 2  n e 2   A 
the increase in resistance  i.e., R1 - R 0 
i 
is directly proportional to the initial resistance, i.e.,
R1  R0  R0
m
Resistivity of a material,  = 2
ne 

Since m and
e are constanct,
 
1
n
Thus, resistivity of a material depends upon electron
density n ( i.e., number of free electrons per unit volume)
and relaxation time

  .
Relaxation time decreases with the increase in
temperature and hence the resistivity increases.
Metals: In most of the metals, the value of n dose not
change with temperature so that

1

 ii 
is directly proportional to the rise in temperature, i.e.,
 iii 
depends upon the nature of the materal.
R1  R0  t1
Combining the first two, we get,
R1  R0  R0 t1
R1  R0   R0 t1
or
where  is a constance of proportionality and is called
temperature co-efficient of resistance. Its value depecds
upon the nature of the material and temperature.
Rearranging eq.
 i  , we get, R 1
Definition of  : From eq.
= R 0 1 +  t1 
 i  , we get 
=
R 1  R0
R0  t1
 increase in resistance/ohm oiginal resistance/ C
rise in temperature
Hence, temperature co-efficient of resistance of a conductor
2.17. NON-OHMIC CONDUCTORS
Those conductors which do not obey
ohm's law  I  V 
are called non-ohm's conductor.
e.g., vacuum tubes, transistors,
electrolytes, etc. A non-ohmic condutor
may have one or more
of the following proporties:
is the increase in resistance per ohm original resistance per C
rise in temperature.
2.18. RESISTORS IN SERIES
A number of resistors are said to be connected in series if the
same current flows through each resistor and there is only one
path for the current flow throughout
 i  The V-I graph is non-linear.
 ii  The V-I graph may not pass through
the origin as in case of an ohmic conductor.
 iii  A non-ohmic conductor may conduct
poorly or not at all
when the p.d. is reversed.
The non-liner circuit problems are generally solved
by graphical methods
Consider three resistors of resistances R, R and R connected
in series across a battery of e.m.f. E volts as shown in.Let I
be the circuit current.By ohm's,
V1  I R1 ; V2  I R2 ; V3  I R3
Now,
E  V1  V2  V3  IR1  IR 2  IR3
or
E  I  R1  R 2  R3 
or
E I  R1  R2  R3
But E I is the total or equivalent resistance R s between point A and B

R s  R1  R2  R3
28
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Hence, when a number of resistances are connected in series, the total resistance is equal to the sum of the individual resistances
.
2.19. RESISTORS IN PARALLEL
NOTE: (i) When the cell is delivering no current the p.d.
across the terminals of the cell is equal to e.m.f. E of the
cell as shown in fig.
 ii  When a resistance R is connected across the cell current I
starts flowing in the circuit. This current causes a voltage drop
 I r 
across internal resistance of the cell so that terminal
voltage V is less than the e.m.f. E of the cell
Internal resistance of a cell:
InternalResistance of a cell:
Now
or
or
E
E
E
I1 
; I2 
; I3 
;
R1
R2
R3
E
E
E
I  I1  I 2  I 3 


R 1 R 2 R3
1
1
1 
I E

 
 R 1 R2 R3 
I
I
I
I



E R1 R 2 R 3
E
Rr
or
IR  Ir  E
But I R  V  Terminal p.d. of the cell.

V  Ir  E
or
E  V  Ir
I 
Internal resistance of the cell,
r
E  V E V 

R
I
 V 
But E I is the total or equivalent resistance R of the
parallel connected resistors so that I E  I R

3.1. CELL IN SERIES
I
I
I
I
  
R R1 R2 R3
2.20. E.M.F. AND TERMINAL P.D. OF A CELL
Total battery e.m.f.  nE
Internal resistance of the battery  nr
Total circuit resistance  R  nr
nE

Circuit current, I 
R  nr
Special cases : (i ) If R  nr then nr can be neglected
as compared to R.
E
 I=n  n  current due to one cell.
R
(i ) If R  nr then R can be neglected
as compared to nr.
nE E
 I=
  current due to one cell.
nr r
29
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Hence in order to get maximum current in series grouping
of cells, the external resistance must be smaller than the
internal resistance of the cell.
3.2. CELLS IN PARALLEL
,(2) Kirchhoff‟s Second Law( The Loop Law or Kirchhoff‟s
voltage law)
It states that the algebraic sum of the products of the
currents and resistances of any closed part( loop ) of an
electrical circuit is equal to the algebraic sum of the
e.m.f.'s acting in that part(loop) of the circuit.
i.e,
 IR=  E
3.4. Wheatstone Bridge
Wheatstone bridge is an arrangement of four resistances used for
measuring one unknown resistance in terms of other three known
resistances.
EMF of thebattery=E
Since the cells are connected in parallel their
internal resistance are also in parallel. If rp is the total
resistance of the battery then,
1 1 1 1
= + + +........m terms.
rp r r r
1+1+1+..............m terms m
=
r
r
r
 rp =
m
=
r
m
E
mE
 Circuit current , I=
=
r
mR+r
R+
m
mE
I=
mR+r
Special cases:
(i) If R<<r, then mR may be neglected as compared
Total circuit resistance =R+rp =R+
to r.
E
=m×current due to one cell.
r
(ii) If r<<R, then r may be neglected as compared
to mR.
mE E
 I=
= =current due to one cell.
mR R
 I=m
Principle;
P R
i.e,

Q S
Q
S  R
P
Applying Kirchhoff's loop law to the closed loop
ADBA, we get
or ,
I1 R  I g G  I 2 P  0, where G is the resistance of the
galvanometer.       (1)
Applying Kirchhoff's loop law to the closed loop
BDCB, we get
I g G  ( I1  I g ) S  ( I 2  I g )Q  0
      (2)
In the case of balanced Wheatstone bridge, no
3.3. Kirchhoff‟s Laws
current flows through the galvanometer
i.e, I g =0
(1) Kirchhoff‟s First Law ( The Junction Law or
Kirchhoff‟s Current Law)
It states that the sum of all the currents entering any
point (junction) must be equal to the sum of all currents
leaving that point(junction).
or,
Hence equation (1) becomes
The algebraic sum of all the currents meeting at a
point ( junction) in a closed electrical circuit is zero.
i.e,
 I=0
Consider a point or junction O in an electrical
circuit. Let I1 and I3 be the currents entering the point
O;I 2 , I 4 and I5 be the currents leaving the point O.
I1 R  I 2 P  0
or ,
or , I1 R  I 2 P
I1 P

     (3)
I2 R
similarly equation (2) becomes,
I1S  I 2Q  0
or , I1S  I 2Q
or ,
I1 Q

     (4)
I2
S
from equation (3) & (4)
P Q

R S
or ,
P R

Q S
Then according to the first law,
I1  I3  I 2  I 4  I5
3.5. Metre Bridge (Slide Wire Bridge)
30
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Principle:
It works on the principle that potential difference
across any part of a uniform wire is directly proportional
to the length of that portion when a constant current
flows through the wire.
Let V be the potential difference across the portion
of wire of length l , resistance R, and uniform area of
cross- section A. If I be the current flowing through the wire
then according to Ohm's law
V=IR
It is a practical or refined form of Wheatstone
bridge. It works on the principle of Wheatstone
bridge.
P R
Q
=
or, S=  R
Q S
P
Working:
Let the resistance of the wire between A and J = P
l
A
where  is the resistivity of the wire.
But,
R=

V= I 
i.e,
and the resistance of the wire between B and J = Q
If r be the resistance per unit length of wire, then
P = rl and Q = r ( 100  l )
According to the principle of Wheatstone bridge
But ,

l I

l
A  A 
I
 a constant, say K
A
V  Kl
or , V  l , provided A,  and I are constant
which is the principle of the potentiometer.
Q
or, S=   R
P
r ( 100  l )
 100  l 

S=
 R or, S= 
R
rl
l


Simply by knowing the values of l and R, we can
P R

Q S
calculate the unknown resistance (S).
3.6. Potentiometer
A potentiometer is a device commonly used for comparing
e.m.f.‟s and to measure internal resistance of cells.
Uses / Applications of Potentiometer
Potentiometer cab be used mainly to
(1) compare the e.m.f.‘s of two cells.
(2) determine the internal resistance of a cell.
POWER QUESTIONS
Q1. A wire is carrying current. Is it electrically
charged?
A. No. The current in a wire is due to flow of free electrons
in a definite direction. The number of protons in the wire at
any instant is equal to the number of electrons and charge
on electron is equal and opposite to that of proton.
Q 2. The flow of charged particles in a definite direction
constitutes an electric current. Even then the current is
a scalar quantity. Explain why ?
Ans. Although electric current has magnitude as well as
direction yet it is a scalar quantity because the law of vector
addition is not applicable to currents as the currents can be
added algebraically.
Q 3. Does drift velocity changes with respect to the
magnitude of current through the conductor.
A. Yes. Drift velocity is directly proportional to the current.
Q 4. The electric force should cause acceleration to the
electrons inside the conductor, then why do the
electrons acquire a steady mean drift velocity ?
Ans. The force acting on the electron due to electric field
accelerates the electron. Hence velocity of the electron
increases which becomes disrupted when the electron
collides with the positive ions of the metal. After collision,
the electron is once again accelerated till it collides with
ions again. So basically, the motion of electrons in the
conductor under the electric field is stop and go motion. In
the absence of continuous acceleration, the electrons
appear to drift with a certain steady velocity.
31
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Q 5. The electron drift velocity is very small and
electron charge is also very-very small . Even then, how
can we set up large current in the conductors ?
Ans. The number density of free electrons (i.e. number of
free electrons/volume) is very large. Due these large
number of electrons large current is produced in the
conductor.
Q6. Is Ohm‟s law a universal law ? Support your
answer with examples.
Ans. No, Ohm‘s law is valid in case of most of metallic
conductors and that also when physical conditions like
temperature, stress etc. of the conductor remains the same.
This law fails in case of vacuum tubes, semiconductor
diodes, thermistors, thyristor etc.
Q 7. Manganin or Eureka is used for making standard
resistance coils. Why ?
Ans. Manganin or Eureka has very high resistivity.
Therefore, a smaller length of the wire of such a material is
required to prepare the coil of given resistance. Also, the
variation of resistance of such materials with temperature is
very small. That is why they are used for making standard
resistance coils.
Q 8. Out of metals and alloys, which has higher value
of temperature coefficient of resistance ?
Ans. Metals have higher value of temperature coefficient
of resistance than the alloys.
Q 9. Why resistance of a super conductor becomes
almost zero ?
Ans. The free electrons in super-conductors are mutually
coherent and co-operative at critical temperature and hence
they do not collide with each other and ions of the
conductor.
So super-conductors have almost zero
resistances.
Q 10. What should be the properties of the standard
resistances ?
Ans. The properties of standard resistances should be :
(i) Its value should not change with time.
(ii) It should show negligible variation with temperature.
(iii) It should have proper capacity for carrying current
without over heating.
(iv) It should have low inductance and self capacitance.
Q 11. When we switch on an electric bulb, it lights
almost instantaneously though drift velocity of electron
in copper wires is very small. Explain.
Ans. When we switch on the bulb, the electric circuit gets
closed, the electric field is established in the circuit
instantaneously with the speed of electro-magnetic waves
causing at every point a local electron drift. The
establishment of current does not wait for the electron to
flow from one end of circuit to another end. Hence the bulb
lights almost instantaneously.
Q 12. Is Ohm‟s law true for all conductors ?
Ans. No, it is true only for metallic conductors.
Q 13. What is the effect of temperature on the
relaxation time of electrons in a metal ?
Ans. Relaxation time decreases with increase in
temperature.
Q 14. What do you mean by a linear resistor ?
Ans. A linear resistor is one which obeys Ohm‘s law i.e. for
which voltage current graph is a straight line.
Q 15. What is the effect of rise in temperature on the
electric resistivity of semiconductor.
Ans. Electrical resistivity of semi-conductor decreases with
rise of temperature.
Q 16. Name three materials whose resistivity decreases
with rise of temperature.
Ans. Silicon, germanium and carbon.
Q 17. If the radius of the copper wire is doubled, what
will be the effect on its specific resistance ?
Ans. Specific resistance remains unchanged.
Q 18. What is the internal resistance of a cell due to ?
Ans. Internal resistance of a cell depends upon : (i) the
nature of electrolyte, (ii) the nature of electrodes, (iii) the
distance between the electrodes and (iv) area of the
electrodes immersed in the electrolyte.
Q 19.Will the drift speed of free electrons in a metallic
conductor increase or decrease with the increase in
temperature ?
Ans. With the increase in temperature, resistance increases
and hence drift velocity decreases.
Q 20. A wire of resistance 4R is bent in the form of a
circle. What is the effective resistance between the ends
of diameter ?
Ans. R.
Q21. When cells are connected in parallel, what will be
the effect on (i) current capacity (ii) e,m.f of the cells.
Ans. (i) Current capacity increases (ii) The effective e.m.f.
of the cells in parallel will be equal to e.m.f. of one cell.
Q 22. How is the current conducted in metals ?
Explain.
Ans. Every metal conductor has large number of free
electrons which move at random at room temperature.
Their average thermal velocity at any instant is zero. When
a pot. diff. is applied across the ends of a conductor, an
electric field is set up. Due to which the free electrons of
the conductor experience force due to electric field and
drift towards the positive end of the conductor, causing the
electric current (i.e. conduction current) in the conductor
whose direction is opposite to the direction of motion of the
free electrons in the conductor.
Q 23. Define ampere, volt and ohm.
Ans. If 1 coulomb charge is flowing per second through a
cross-section of a conductor, then current through the
conductor is 1 ampere.
If 1 joule work is done in bringing 1 coulomb of charge
from one point to another against the electric forces, then
pot. diff. between those two points is 1 volt.
If 1 ampere current flows through a conductor when 1 volt
pot. diff. is applied across its two ends, then the resistance
of conductor is 1 ohm.
32
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Q 24. What is terminal potential difference of a cell ?
Can its value be greater than the e.m.f. of a cell?
Explain.
Ans. Terminal potential difference of a cell is defined as
the potential difference between the two electrodes of a cell
in a closed circuit. The value of terminal potential
difference of a cell is less than the e.m.f. of a cell, when
current is drawn from the cell (i.e. during discharging of
cell). The value of terminal potential difference of a cell
becomes greater than the e.m.f. of the cell during charging
of the cell i.e. the positive electrode of the cell is connected
to positive terminal of battery charger and negative
electrode of the cell is connected to negative terminal of
battery charger.
Q 27. What is super-conductivity ? Write its two
applications.
Ans. Superconductivity is a phenomenon of disappearance
of all signs of resistance in a conductor at a critical
temperature or transition temperature which is usually low.
Superconductivity can be used (i) to produce very high
speed computers and (ii} to transmit electricaj power from
generating station to consumers over supper conducting
cables without any loss of power.
Q 28. What are super-conductors ? Write its two
applications.
Ans. As the temperature of certain metals and alloys
decreases, their resistance also decreases. When the
temperature reaches a certain critical value called critical
temperature, the resistance of material completely
disappears i.e. it becomes zero. Then the material behaves
as a super- conductor.
Thus super-conductors are those material conductors
whose resistance disappear at critical temperature. The
critical temperature is different for different materials.
Super conductors are used (i) in power transmission (ii) to
produce very high speed computers.
Q29. Draw V-I graph for ohmic and non-ohmic
materials. Give one example for each.
Ans. V-I graph for an ohmic material is a straight line
passing through origin. . Example is a resistor made of
manganin.
V-I graph for a non-ohmic material is a curve , non-linear
or straight line not passing through origin. Example is
electrolyte or junction diode.
Q35. Define resistance of a conductor. What is its
cause?
Explain
the
factors on which the resistance of a conductor depends.
Ans. The resistance of a conductor is the obstruction posed
by the conductor to the flow of electric current through it.
Resistance of a given conducting wire is due to the
collisions of free electrons
with the ions or atoms of the conductor while drifting
towards the positive end of the conductor
The resistance of a conductor depends upon the following
factors :
Q25. State the factors on which (i) internal resistance
(it) e.m.f. of a cell depend.
Ans. Internal resistance of a cell depends upon ; (i) distance
between the plates (ii) the nature of electrolyte (iii) the
nature of electrodes (iv) area of the plates, immersed in the
electrolyte. If area increases, internal resistance decreases.
E.M.F. of a cell depends upon; (i) nature of electrodes (ii)
nature and concentration of electrolyte used in the cell (iii)
temperature of electrolyte.
Q 26. What is the e.m.f of a cell ? On what factors does
it depend ?
Ans. E.M.E of a cell is defined as the maximum value of
potential difference between the two electrodes of a cell
when the cell in the open circuit i.e. no current is drawn
from the cell.
E.M.F of a cell depends upon (i) nature of electrodes (ii)
nature and concentration of electrolyte and (iii) temperature
of electrolyte used in cell.
(i)
The resistance (R) of a conductor is directly
proportional to the length of the conductor(ii)
The
(ii)
resistance of a conductor is inversely
proportional to the area of cross-section of the
conductor
The resistance of a conductor also depends upon the nature
of material and temperature of the conductor.
Q30. Define the term „resistivity‟ and „conductivity‟
and state their S.I. units. Draw a graph showing the
variatia of resistivity with temperature for a typical
semiconductor.
A. Resistance offered by a unit cube(unit length and and
unit area of cross-section) of a conductor is called
resistivity. Its unit is ohm-metre.
The inverse of resistiviiy of a conductor is called Its
electrical conductivity
The unit of electrical conductivity is mho m-1 or Sm-1
Q 31. Explain, why bending a wire does not affect
electrical resistance ?
Ans. Free electrons in a wire have small value of drift
velocity and hence low value of inertia of motion. Due to it,
they are able to go around the bends easily.
Q 32. Are Kirchhoffs laws applicable to both a.c. and
d.c. ?
A. Yes, Kirchhoff‘s laws are equally applicable to a.c. as
well as d.c. circuits.
Q 33. Why do we prefer a potentiometer to measure
e.m.f. of a cell rather than a voltmeter?
A. A potentiometer does not draw any current from the cell
whose e.m.f, is to be determined, whereas a voltmeter
always draws some current. Therefore, e.m.f. measured by
voltmeter is slightly less than actual value of e.m.f. of the
cell.
Q 34.The e.m.f, of the driver cell in the potentiometer
experiment should be greater than the e.m.f. of the cell
to be determined. Why?
A. If it is not so, there will be smaller fall of potential
across the potentiometer wire than the e.m.f. of the cell to
be determined and hence the balance point will not be
obtained on the potentiometer wire.
33
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Q 37.Why do we prefer potentiometer of longer length
for accurate measurements ?
Sol. The longer is the length of a potentiometer wire, the
more accurate is the measurement with it. It results in lesser
value of fall of potential per unit length of wire i.e. lesser
value of least count.
Q38. Copper wire is not used in potentiometers. Why ?
Sol. It is so because the temperature coefficient of
resistance of copper is large and its resistivity is small.
Q 123.On what factors, does the potential gradient of
the potentiometer wire depend ?
Q39. Can Metre bridge be used for finding the
resistance of (i) moderate values (ii) high values (iii) low
values ? Explain.
A. The metre-bridge is most sensitive when the resistance
of all the four arms of the bridge are of the same order. It
is so for moderate resistances only. The Metre bridge
becomes insensitive for too high or too low resistances.
While measuring low resistances, the resistances of the
copper strips and connecting wires become comparable to
the unknown low resistance and hence cannot be neglected.
Q 40. When is Wheatstone Bridge most sensitive ?
A. Wheatstone Bridge is most sensitive when the value of
resistance of the four arms of the bridge is of the same
order. It is due to this reason, the balance point is obtained
at the middle point of metre bridge wire.
Q41.What is potential gradient ? State its S.I. unit.
Ans. Potential gradient is defined as the fall of potential
per unit length of potentiometer wire. Its unit is Vm -1.
Q42. If the length of the wire be (i) doubled and (ii)
halved, what will be the effect on the position of zero
deflection in a potentiometer ? Explain.
Ans. (i) When length of the wire is doubled. The potential
gradient across the potentiometer wire will decrease. Due
to which the position of zero deflection will occur at longer
length, (ii) The reverse will be the case as first.
Q 43. How can you make a potentiometer of a given
length more sensitive by using a resistance box?
Ans. The sensitivity of a potentiometer is the smallest
potential difference it can measure. It can be increased by
reducing the potential gradient i.e. potential drop per unit
length of potentiometer wire. The same is possible by
decreasing the current flowing in the potentiometer wire.
Q 44. Kirchhoff‟s law obeys the law of conservation of
charge. Explain.
A. According to Kirchhoff‘s law, the current entering a
junction is equal to the current leaving the junction in a
certain time. This shows that Kirchhoff‘s law obeys the law
of conservation of charge.
Q45. A wire connected to a bulb do not glow, whereas
the filament of the bulb glows when same current flows
through them. Why ?
Sol. Filament of bulb and supply wires are connected in
series so the same current flows through them. Since the
resistance of connecting wires is negligibly small as
compared to the resistance of filament and heat produced
due to given current is directly proportional to its resistance
(from Joule‟s law of heating), therefore, the heat produced
in the filament is vey large. Hence the bulb glows.
Q46. Explain why an electric bulb becomes dim when
an electric heater in parallel circuit is made on. Why
dimness decreases after some time ?
Sol. Since the electric heater has more power than that of
electric bulb and power is reciprocal to resistance for a
given supply voltage, hence the resistance of heater coil is
less than that of electric bulb.
Q47. Nichrome and copper wires of same length and
same diameter are connected one by one between two
points of constant potential difference. In which wire
the heat will be produced at higher rate ? Explain.
Sol. Since the resistivity of nichrome is more than copper,
therefore, the resistance of copper wire is less than that of
nichrome wire, for the given length and diameter. As rate
of heat production, P = V 2/R . Therefore, in copper wire
rate of heat production is mors.
Q48. How does the use of fuse wire save the electrical
installations ?
Sol. A fuse wire is one which has high resistance and low
melting point. This is connected in series with electrical
installations. When the supply voltage exceeds the safe
limit, more heat is produced in the fuse wire. ( as rate of
heat production= V2/R), which melts it. Due to which the
circuit breaks and the damage to the electrical installations
is saved.
Q49. What is the law that defines heat produced by an
electric current ?
Ans. Joule law of heating. It states that the amount of heat
produced in a conductor when a constant current flows
through it is directly proportional to (i) the square of the
current, (ii) the resistance of the conductor and (iii) the time
for which current is passed.
Q 50. Why is the conductivity of an electrolyte low as
compared to a metal?
Ans. The current in an electrolyte is due to the drifting of
ions (+ve as well as negative) whereas in a metallic
conductor, it is due to the drifting of free electrons.
The conductivity of an electrolyte is very low as compared
to the metallic conductors on account of the following
reasons ; (i)
The number density (n) of ions in an
electrolyte is very small as compared to that of free
electrons in a metallic conductor.
(II) Since the mass of ions is very large as compared to that
of electrons, the drift velocity of ions is smaller than that of
electrons in a given electric field.
(iii) The resistance offered by the solution to ions is much
more than that offered by the metal to the drifting of free
electrons.
Q 51. Can we use a.c. for electrolysis?
Ans. No. Electrolysis is possible only if d.c. potential
difference is applied to the electrodes. It is because we are
to attract ions of only one kind on each electrode.
Q 52. Why is the use of Leclanche cell preferred in
metre-bridge experiments?
34
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Ans. In meter-bridge experiments, we require low and
discontinuous current. For this reason, the use of
Leclanche cell is preferred in these experiments.
Q 53. What are the advantages of secondary cells over
the primary cells?
Ans. The following are the advantages of secondary cells
over the primary cells : (i) The e.m.f. of a secondary cell
is high. (ii) A secondary cell can be recharged. (iii) A
secondary cell can deliver steady current. (iv) The internal
resistance of a secondary cell is small, (v) The capacity of
a secondary cell is large.
UNIT III (Magnetic Effects of Current)
In 1820, Danish scientist Oersted discovered that a magnetic field is associated with a conductor carrying current. Oersted
observed that when electric current passes through a conductor, then magnetic field is produced around it. If a strong electric
current is passed through the conductor, then magnetic field produced around the conductor is also very strong and the earth‘s
magnetic field may be cancelled. The direction of the magnetic field may be found using the right hand thumb rule or Maxwell‘s
cork screw rule.
3.7. Laplace‟s law (or Biot Savarat‟s law)
Laplace‘s law is used to find the magnetic field at a point near a conductor carrying current. This law was first experimentally
confirmed by Biot and Savart and for this reason it is also known as Biot Savarts‘s law.
According to Biot Savart law, the magnetic field dB at a point P due to a small elementary portion dl of the conductor carrying
current is
(i)
(ii)
(iii)
(iv)
(v)
directly proportional to the current i.e. dB  I
directly proportional to the length of the elementary portion i.e. dB  dl
directly proportional to the angle

between the direction of flow of current and the line joining the elementary
portion to the observation point i.e. dB  sin 
inversely proportional to the square of the distance of the point from the current element i.e.
dB 
1
r2
35
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Mathematically,
 B   dB sin 
Idl sin 
r2
 Idl sin 

 dB  0
; where 0  107 WbA1m 1
4
r2
4
 107 TA1m
From Biot sa var t ' s law, dB 
dB 
1T  1Wbm 2 
 
 B. dl   I
0 Idl sin 
4
r2
0 Idl
  900 
2 
4

r
Proof: Consider a circle of radius a.Let XY be a small element of


0 Idl
 dl
B .dland
dBBsin
sin  because
length
areinthe
direction
 4same
 r2

I to the circle.
direction of B is along thetangent
 0 2 sin   dl
 
4 r
 B . dl  Bdl cos   Bdl cos 00  Bdl.
But  dl
 2 a (circumference of loop )
Taking line
integral over the closed path
0

0 I 
sinB2dl a
B. dl
4 r 2
  21 
substituting the valueaof B from equation, B= 0  
Since sin  
4  a 
r
to the right side, we get
0 I a
0 2 I a 2
B 

2

a


I
21
0
4B. dlr 2 r 4 a dl 420 ar3 dl
Since r 2  a 2  x 2
B 
In S.I. unit, the magnetic field dB is measured in tesla
3.8. Magnetic field due to a circular loop
Let P be a point on the axis of the circular loop. The components of
magnetic field due to element dl dBcos  (vertically upward and
downward) cancel each other as they act opposite along the same line.
Therefore, effective component of magnetic field for the current
element=dBsinϴ(along axis)
 dl  circumference1 of the circle=2 a
But ,
r
B. dl
 a 
 2x a

2 a 2
0 I2
 


B  0
4
2 2
 
or,
2 Ia
a
 B. dl   I
0
3
2
 x2  2
If the coil has n turns ,
B
Therefore, the magnetic field due to the whole loop can be found by
integrating the magnetic field produced by the small current element
dBsin
a
x

Special cases: (i) If the point P is at the
centre
of the loop,
 Ampere’s
lawx=0
holds good for a closed path
of any shape and size.
0 2Biot
 nIsavart law and Ampere’s circuital

 B Both
4

a
law
give same physical results of electric
(ii) If observation point P lies far away from
current but in two
different way
the coil, x>>a, then a2 can be neglected as
compared to x2.
Solenoid: A cylindrical coil of many tightly wound
turns of insulated
wire
with generally diameter of the
2
0 2 nIa
 B  than
coil smaller
its
length
is called a Solenoid.
4 x3
.
3.9. Current loop as magnetic dipole
For a magnetic dipole, the magnetic field at a point distant x from its
axis is given by B  0 2M
4 x3
The magnetic field due to a circular loop on n turns is given by
B
0 2 nIa 2
3
4 2
2 2
0 2 nIa 2
4 x3
Since A   a 2
B
0 2nIA
4 x3
Comparing the above two equations we get,
M=nIA which is the magnetic dipole moment
of a current loop. Thus a circular loop
produces magnetic field in the same manner as
a magnetic dipole does.
36
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
4.0. Amperes circuital law According to Ampere‘s circuital law the line integral of the magnetic field around any closed path
in free space is equal to absolute permeability of free space (  0 ) times the net current enclosed by the path.
4.1. Magnetic field due to a straight solenoid
Let P be a point well within the solenoid. Consider any rectangular loop ABCD(known as Amperian Loop) passing
through P.
 
 B. dl 
Then,
Line integral of magnetic field across the
loop ABCD
B  
C  
D  
A  
A
B
C
D
=  B. dl   B. dl   B. dl   B. dl
      (1)

B is perpendicular to paths BC and AD i.e. angle between


B and dl is 90 0 for these paths.

C  
A  
B
D
 B. dl   B. dl   Bdl cos90
0
0

since path CD is outside the solenoid where B is taken as 0,
D  
so
 B. dl  0.
C


For path AB, the direction of d l and B is same i.e.  =0
37
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
4.2. Magnetic field due to a torroid carrying current
Hence, equation (1),becomes
 
B  
B
B
A
A
A
 B. dl   B. dl   Bdl cos 0   Bdl
B
 
or ,
 B. dl  B  dl
( B is uniform)
A
 
or ,
 B. dl  Bl
 B

  dl  total length of pathAB=l     (2)
 A

According to Ampere's circuital law
 
 B. dl  
0
 net current enclosed by loop ABCD.
= 0  number of turns in the loop ABCD  I
 0 nlI  (3)
Comparing equation (2)&(3),we get
Bl  0 nlI
B  0 nI
Thus, magnetic field well within an infinitely long solenoid is
given by,
B  0 nI .
A ring shaped closed solenoid is known as torroid.
In a torroid, the magnetic field lines are in the form
of concentric circles.
4.3. Force on a moving charge in a magnetic field
(magnetic Lorentz force)
 
 B. dl   Bdl cos 


The magnetic field B is tan gential to the element dl , so
angle between them  0.

 
 B. dl   Bdl cos 0  Bdl
 B  dl  B  circumference of circle of radius r.
 
  B. dl  B  2 r  (1)
From Ampere ' s circuital law,
 
 B. dl  
The force on a charge moving in a magnetic field is
i.e. F  B
From (1) and (2),
B  2 r  0  (n  2 r ) I
 B  0 nI
Special cases:
(i)
(ii) directly proportional to the magnitude of charge. i.e.
F q
(iii) directly proportional to the component of velocity
along a direction perpendicular to the direction of magnetic
field. i.e. F  v sin 
(ii)
Combining the three factors,
F  kBqv sin 
The cons tan t of proportionality k  1( SI )
 F  Bqv sin 
 total current through torroid
 0  (n  2 r ) I  (2)
called magnetic Lorentz force. The magnitude of the force
acting on the charge is
(i) directly proportional to the magnitude of the strength of
magnetic field
0
(iii)
If the charged particle is at rest. Here, v=0.
Therefore, the equation simplifies to F=qB(0)sin
 =0. Hence, a charged particle at rest in a
magnetic field does not experience any force.
(ii) When the charged particle moves parallel to
magnetic field,  =0 0r π, therefore,
F=qvBsin0=0. So, if a charge particle moves
parallel or antiparallel to magnetic field, it does
not experience any force.
(iii) When the charge particle moves
perpendicular to the magnetic field,  =900,
therefore, F=qvBsin900=qvB. (maximum force)
38
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
4.8. Moving Coil Galvanometer
4.4. Torque on a Current Loop in a Magnetic field.
Moving Coil Galvanometer is a device used to detect /
measure small electric current flowing in the electric circuit.
Principle: Moving coil galvanometer is based on the fact that
when a current carrying loop or coil is placed in the uniform
magnetic field, it experience a torque.
Theory
Let B  Intensity of magnetic field
I  Current flowing through the coil
l  Length of coil
b  Breadth of the coil
(l  b)  A  Area of the coil


F 1 and F 2 form a couple and try to rotate the loop anticlockwise
The magnitude of the torque ( ) due to forces

Then,  N  NIAB
This torque is known as the deflecting torque.
 
 I  l  B   DN


Restoring torque is given by
 I  lB sin 900   b sin 
  I  lb  B sin 
 ' N  k
For equilibrium of the coil, Deflecting torque = Restoring torque
i.e, NIAB = k
Since,  lb   A, area of the loop

  IAB sin 
If the loop has N turns, then net torque acting on the loop is
 N  N  NIAB sin 
 k 
I 

 NAB 
or ,
I  G
k
where
G
and is called galvanometer constant.
NAB
or,
I 
Thus, deflection of the coil is directly proportional to the current flowing
or ,
Another consideration,

If the plane of loop makes an angle  ' with the magnetic field B then
   '=900 or ,    900   ' 
Hence, the equation becomes
 N  NIAB sin  900   ' 
or ,
 N  NIAB sin

F 1 and F 2 is given by
 =Magnitude of the either force  lever arm =F1  DN
or ,
N  Number of turns in the coil
when current flows through the coil, it experiences a torque, which is given by
through it. Hence we can use a linear scale in the galvanometer to detect the
current in the circuit.
 N  NIAB cos  '
4.9. Sensitivity of a galvanometer: A galvanometer is
Unit of strength of magnetic field. In SI unit, the unit of
strength of magnetic field is tesla(T).
F  qvB sin 
To define one tesla
F
B
qv sin 
1tesla (T ) 
1N
 1NA1m 1
1C 1ms 1 1
Hence, the strength of magnetic field at any point is called tesla
if a charge of one coulomb when moving with a velocity of 1ms-1
along a direction perpendicular to the direction of magnetic
field, experiences a force of one newton
said to be sensitive if a small current flowing through the coil
of galvanometer produces a large deflection in it.
Current sensitivity:The current sensitivity of a galvanometer
is defined as the deflection produced in the galvanometer per
unit current flowing through it.
i.e, Current sensitivity 




   NAB  NAB


I
k
k
k 

 I 

NAB


1tesla (T) = 1 weber metre-2( Wb m-2)
( unit of magnetic flux is weber-Wb)
1T = 1NA-1m-1 = (1Wb m-2)
Strength of magnetic field is commonly
expressed as gauss (G)
1 gauss(G) = 10-4 tesla (T)
If the charged particle is located in electric
field,it will experience a force whether the
body is at rest or not.
39
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
4.6. Lorentz Force due to electric and magnetic
fields
Conversion of galvanometer into Ammeter and
Voltmeter:
When a charged particle having charge q moves

in a region where both electric field E and magnetic

field B exist, it experiences a net force called
Lorentz force(F).

Lorentz ( F)  Force on charge due to electric
field + Force on charge due to
magnetic field


= Fe  Fm
or ,




 
 
F  q E  q  v  B  or , F  q[E   v  B  ]




Let G and S be the resistance of a galvanometer and shunt respectively.Let I be the
total current to be measured by an ammeter in the circuit.
Definition of Magnetic Field B from Lorentz Force
Lorentz magnetic force is given by Fm  qvB sin 
Fm
qv sin 
where q is the charge on a particle, v is the velocity of the particle in
B
or ,
magnetic field B.
If
q  1, v  1 and   900 , then
B  Fm
Therefore, Strength of magnetic field (B) at a point may be
defined as the magnetic force experienced by a unit charge
moving with unit velocity at right angles to the magnetic
field.
4.7. Expression for force between two infinitely
long straight parallel conductors
Let Ig be the current flowing through the galvanometer corresponding to which
galvanometer gives the full scale deflection.The remaining current (I-I g ) is to flow
through the shunt.
Since G and S are parallel, the potential difference across them is same.
 Ig 
or S    G
 I-I 
 g
This is the required value of shunt resistance to convert a galvanometer into an ammeter
IgG  (I-Ig ) S
i.e,
of range 0 - I ampere.
4.11. Effective resistance of ammeter
The total effective resistance is given by
1
1 1 GS
GS
=  
or , R eff 
R eff G S
GS
GS
G  S , G  S  G ( S  negligible)
GS
Hence, R eff 
S
G
Since
4.12. Voltmeter A voltmeter is an instrument used to
measure the potential difference across the two ends of a circuit
element.
Conversion of Galvanometer into VoltmeterA
galvanometer can be converted into a voltmeter by connecting a
large resistance in series to the galvanometer. Let G and R be the
resistance of a galvanometer and a conductor connected in series
with it respectively. Let V volt be the potential difference to be
measured by the voltmeter.
From Biot-Savart law, expression for magnetic field
intensity at a point due to current carrying conductor is
Let Ig be the current flowing in the circuit corresponding to which the voltmeter
gives the full scale deflection.
  2I 
B 0 
 (1)

4  r 
  2 I1 
BX  0 

4  r 
0  2 I 2 
BY 


4  r 
Then, F2  BX I 2l
 force
Now potential difference between points A and B is mgiven by
V=Ig R  IgG  Ig  R  G 

per unit length; l  1
 F2  BX I 2
0  2 I1 

 I2
4  r 
0  2 I 2 
Similarly , F1 

 I1
4  r 
Forces F1 & F2 are equal in magnitude.

or ,
RG 
R
V
Ig
V
G
Ig
This is the required value of resistance which must be connected in series to the
galvanometer to convert it into a voltmeter of range 0 - V volt.
Effective Resistance of the Voltmeter Effective value of
resistance of the voltmeter is given by R‘ = ( R + G ), which is
very high.Thus, voltmeter is a high resistance device. Resistance
of an ideal voltmeter is Infinite.
40
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)

From Fleming‟s left hand rule, if currents
I1 and I2 have direction, they attract. If the
(2).Magnetic field intensity at a point on EQUATORIAL
LINE of bar magnet.
currents are in opposite direction, they
repel(explanation of Art.4.7)
Current sensitivity of galvanometer can be
increased either by,
(a) increasing the magnetic field B by using a strong
permanent horse-shoe shaped magnet.
(b) increasing the number of turns N. ( But, number
of turns of the coil can not be increased beyond a
certain limit. This is because the resistance of the
galvanometer will increase subsequently and hence
the galvanometer becomes less sensitive).
(c) increasing the area of the coil A. ( But this will
make the galvanometer bulky and ultimately less
sensitive).
(d) decreasing the value of restoring force constant k
by using a flat strip of phosphor-bronze instead of a
circular wire of phosphor- bronze because the value
of k is small in case of a flat strip than a round wire
of phosphor bronze.

B1 

4.10. Advantages of a moving coil
galvanometer
(1) It can be made extremely sensitive, so even a
minute current in the electric circuit can be detected.
(2) Since deflection of the coil of the galvanometer
is directly proportional to the current, so a linear
scale can be used.
(3) As the magnetic field (B) is very strong, so the
external magnetic fields (say horizontal component
of earth‘s magnetic field) can not change the
deflection of the coil of the galvanometer. Thus, the
galvanometer can be set in any position.
(4) It is dead beat type galvanometer. ( Dead beat
type galvanometer comes to the rest position quickly
after disturbance, hence the name to this type of
galvanometer is given as dead beat galvanometer)
When the coil rotates in the magnetic field, eddy
currents are produced in the matallic frame on which
the coil is wound. These eddy currents produce a
dampening effect and hence the coil comes to the
position of rest quickly.
(5) Moving coil galvanometer can be made ballistic
by using a non-conducting frame (made of ivory or
bamboo) instead of a metallic frame.
0 m
along NP
4  r 2  l 2 
Magnetic field int ensity at P due to S  pole of thebar magnet ,

B2 

Quartz fibres can also be used for suspension of the
coil because they have large tensile strength and very
low value of k.
0
m
along NP           (1)
4 ( r 2  l 2 )2
0 m
along PS            (2)
4  r 2  l 2 

B1 and B 2 are inclined at angle 2 . Therefore, the resultant of these two field intensities
is given by
Be  B12  B22  2 B1 B2 cos 2
sin ce B1  B2
Be  2 B12  2 B22 cos 2  2 B12 1  cos 2 
(1  cos 2   2cos 2  )
 2 B12  2cos 2 
 2 B1 cos 
u sin g equation (1), we get Be  2 
l
cos  
0

Be 
sin ce
m  2l  M

Be 
4
0
m
cos 
4  r 2  l 2 

r2  l2
m  2l
r
2
 l2 
3/ 2
0
M
4  r 2  l 2 3/ 2
In case magnet is of very small length then l 2   r 2
Be 
0 M
4 r 3
4.13. MAGNETS AND EARTH
Magnetic Dipole and Magnetic Moment
A magnetic dipole consists of a pair of magnetic poles of equal and
opposite strength separated by a small distance. Examples of magnetic
dipoles are magnetic needle, bar magnet, current carrying solenoid, a
current loop etc.Atom is also considered to behave like a dipole so the
fundamental magnetic dipole in nature is associated with the electrons.
41
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
4.5. Force on a current carrying conductor placed
in a magnetic field
Magnetic dipole moment is defined as the product of the pole
strength of either pole and distance between the poles.Distance
between the two poles is called magnetic length and is taken as 2L.
Let m be the pole strength of each pole then magnetic dipole
moment is given by,
M  m  2l
M is a vector quantity so it can be written as



M  m(2l ), where 2l is the magnetic length directed from South to North pole.
Thus direction of magnetic dipole moment is from south to north pole.
S .I .unit of m is ampere- metre (Am) and that of M is ampere - metre2 ( Am2 )
or joule tesla 1 ( JT 1 ).
If the conductor has area of cross section A and the
number of free electrons per unit volume equal to n,
then current flowing through the conductor is given
by
I  enAvd (relation between current and drift velocity )
Multiplying on both sides by l ,
4.14. Magnetic Field due to a Magnetic dipole (Bar magnet)
lI  enAlvd  (1)
force exp erienced by each electron
Magnetic field intensity at a point on the AXIAL LINE of the
bar magnet.
f  evd B sin  (in magnitude)
 f  qvB sin  ; here, q  e
Total number of electrons  n  volume
where n  electron density.
 Total number of electrons  n  Al

Magnetic field intensity at P due to North pole of the bar magnet, B1  Force exp erienced
by unit N  pole at P 
From equation (1)
F  IlB sin 
0 m
along NP
4 (r  l )2

Similarly, magnetic field intensity at P due to South pole of the bar magnet, B2 
along PS .
 Net magnetic field int ensity at P dueto the bar magnet ,
Ba 

0 m
 m
  (r  l ) 2  (r  l ) 2 
 0
 0 m
2
2
4 (r  l ) 4 (r  l ) 4  (r 2  l 2 )2 
0   r  l  r  l  r  l  r  l   0  2r  2l 
m
  m 2 2 2 
4 
(r 2  l 2 )2
 4  (r  l ) 
sin ce, m  2l  M

Ba 
So, total force on conductor  total force on all electrons
 F  (total electron)  force on each electron
 F  (n  Al )  (evd B sin  )
0 m
4 (r  l )2
Fleming‟s Left Hand Rule
It states that if the fore-finger, central finger and the
thumb of left hand are stretched mutually
perpendicular to each other such that the fore-finger
points in the direction of magnetic field (B) and the
central finger in the direction of current (I), then the
thumb points in the direction of the force experienced
(F) by the conductor.
0 2Mr
along NP
4 (r 2  l 2 )2
If the magnet is of very small length then l 2   r 2

Ba 
0 2 M
4 r 3

Direction of B a is along SN extended .
42
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Torque experienced by a magnetic dipole:
4.18. Atom as a Magnetic Dipole and Bohr Magneton
If a magnetic dipole ( bar magnet ) is placed in a uniform
magnetic field then North and South poles of the magnet will
experience equal and opposite forces.
e
T
where, e is the charge on an electron and T is the period of orbital motion.
I=
But
T=


Circumference 
 As v 

Time period 

e
ev

2 r / v 2  r
Magnetic moment of a current loop,
ev
evr
M  IA
 r2 
2 r
2
e
M
L
(orbital motion)
2m

 e 
In vector notation, M   
L
 2m 
Let (1) Magnetic length of magnet = 2l
(2) Pole strength of each pole = m
(3) Strength of magnetic field = B

I
2 r
v

(4) Angle between M and B  


then, Force acting on North pole = mB along B
       (1)

Negative sign shows that M and L are opposite to each other.
According to Bohr's Theory, angular momentum of electrons is given by

Force acting on South pole  mB oppositeto B
These forces constitute a couple which tends to rotate the magnet in the direction of

B, thus the magnet experiences a torque.
 Torque acting on bar magnet is
h
, where n  1, 2,3......and h is Planck ' s cons tan t.
2
e nh
 eh 
Then equation (1) becomes M 
 n

2m 2
 4 m 
L= n
 =Force  distance between forces
eh
, which is Bohr magneton denoted by B . It serves as natural
4 m
unit of magnetic moment.
=mB  ZN=mB  SN sin    mB  2l sin  
If n  1 then M least 
ZN


in  SZN ,sin   or , ZN  SN sin  
SN


  m  2l  B sin  or   MB sin 
Bohr magneton can be defined as the orbital magnetic moment of an electron circulating
in the innermost orbit.
 m  2l  M , the magnetic dipole moment 

In vector form
when

eh 1.6 1019  6.6 1034

4 m 4  3.14  9.11031
 9.27 1024 Am2 .
B 

  M B
B  1unit and   900 then   M
sin90  1
0
Thus, magnetic dipole moment can be defined as the torque acting on a magnetic
dipole placed normaltoa uniformmagneticfieldof unit strength.
4.19. The magnetic elements of earth
Declination (θ): Declination at a place is the angle between the
geographic meridian and the magnetic meridian at that place. It
is denoted by.
Dip or Inclination (δ): Dip or inclination at a place is the angle
between the direction of intensity of earth‟s total
magnetic field (R) and the horizontal direction in the magnetic
meridian at that place.
It is denoted by .
(ii) Horizontal component of Earth‟s magnetic field:
It is the component of earth‟s total magnetic field along
horizontal direction in the magnetic meridian. It is denoted by H
43
[Type a quote from the document or the
summary of an interesting point. You can
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)

Torque acting on a current carrying loop placed in the magnetic field B is given by
  IAB sin         (1)

where, I  Current flowing in the loop, A  Area of the loop, is the angle between B
and normal to the plane the loop.

But, the torque acting on a magnetic dipole placed in the magnetic field B is given by
  MB sin          (2)
Comparison of equation (1) and (2) gives the magnetic dipole moment of current loop
i.e,
4.20. Tangent law in magnetism.
Suppose in the equilibrium position, the dipole moment of the
magnet makes an angle
with the direction of H. Let the pole strength of each pole of the
magnet be m.
Moment of couple due to field H
= m H x SA
... anticlockwise
Moment of couple due to field F
= m F x NA ... clockwise
In the equilibrium position, the moments of the two couples are
equal, m F x NA = m H x SA
or F  H  SA  H tan   F  H tan  which is
NA
tangent law.
The diamagnetic, paramagnetic and ferromagnetic substances
(i) Those substances (e.g. copper, zinc, bismuth etc.) when placed
in a magnetic field, the substance is feebly magnetised in a
direction opposite to that of the applied field is known as
diamagnetic substances.. Therefore, a diamagnetic substance is
feebly repelled by a strong magnet.
(a) When a paramagnetic substance (e.g. aluminium, antimony
etc.) is placed in a magnetic field, the substance is feebly
magnetised in the direction of the applied field. Therefore, a
paramagnetic substance is feebly attracted by a strong magnet.
(ii) When a ferromagnetic substance (e.g. iron, nickel, cobalt etc.)
is placed in a magnetic field, the substance is strongly magnetised
in the direction of the applied field. Therefore, a ferromagnetic
substance is strongly attracted by a magnet.
Note that diamagnetism and paramagnetism are weak forms of
magnetism.
Magnetic dipole moment ( M ) of the current loop.
M = IA
If the loop has n turns, then M  n IA
where n is number of turns of loop, I is the current and A is the area of loop.

In vector form

M nIA
4.21. Important properties of diamagnetic,
paramagnetic and ferromagnetic substances.
(i) A diamagnetic substance is feebly repelled by a strong
magnet.
(ii) When a rod of diamagnetic substance is suspended
freely in a uniform magnetic field, the rod comes to rest
with its axis perpendicular to the direction of the applied
field.
(iii) A diamagnetic substance on being placed in a nonuniform magnetic field begins to move from stronger to
weaker regions of the magnetic field
4.13. Properties of paramagnetic substances.
(i) A paramagnetic substance is feebly attracted by a strong
magnet.
(ii) When a rod of paramagnetic substance is suspended
freely in a uniform magnetic field, the rod comes to rest with
its axis parallel to the applied field.
(iii) In a paramagnetic substance, the external magnetic field
induces a magnetic field that is directed in the same.
direction as the applied field. For this reason, a
paramagnetic substance on being placed in a non-uniform
magnetic field begins to move from weaker to stronger
regions of the magnetic field.
(iv) When a paramagnetic substance is placed in a magnetic
field, it is feebly magnetised in the direction of the applied
field.
44
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
(v) When a paramagnetic substance is placed in a magnetic field, the magnetic field lines prefer to pass through the paramagnetic
substance rather than air.
4.14. Properties of ferromagnetic substances:
(i) A ferromagnetic substance is strongly attracted by a magnet.
(ii) When a rod of ferromagnetic substance is suspended freely in a uniform magnetic field, it aligns itself in the direction of the
applied field.
(iii) When a ferromagnetic substance is placed in a non-uniform magnetic field, it moves from weaker to stronger regions of
magnetic field.
(iv) When a ferromagnetic substance is placed in a magnetic field, it is strongly magnetised in the direction of the applied field.
4.15. Earth‟s Magnetism
Although the cause for the earth‘s magnetism is not exactly known, several reasonable theories have been proposed. We know
that magnetic fields arise from electric currents.
Therefore, all theories explain earth‘s magnetism on the basis of electric currents.
(i) One theory suggests that the magnetism of earth may be due to molten charged metallic fluid in the core of earth. As the earth
rotates about its axis, the charged fluid also rotates. This gives rise to electric currents in the fluid. These electric currents arc
responsible for earth‘s magnetism.
(ii)
Another theory suggests that since earth has charged particles (protons and electrons), the rotation of earth about
its axis generates electric currents. These currents magnetise the earth.
(iii) There is yet another view regarding earth‘s magnetism. High-energy rays coming from the sun collide with the atoms of the
gases in the upper layers of the atmosphere and ionise them.
ADDITIONAL QUESTIONS
Q1. The magnetic field at a point near the centre but outside a current carrying solenoid is zero. Explain why ?
Sol. Consider a vertical plane passing through the centre of the solenoid and perpendicular to its plane, so that it divides the
solenoid into two equal parts. At a point on this plane just outside the solenoid, the magnetic fields produced due to currents in
the two halves of the solenoid are equal in magnitude and opposite in direction. Hence, the resultant magnetic field at such a
point is zero.
Q2. How much force will be experienced by a moving charge in a magnetic field?

 
Sol. F =q (v x B)
Q 3. Where is the magnetic field due to current through a circular loop uniform ?
Ans. At the centre of current loop.
Q 4. What is the nature of magnetic lines of force due to current in a straight conductor ?
Ans. Magnetic lines of force are in the form of concentric circles with the conductor as centre, lying in a plane perpendicular to
the straight conductor.
Q 5. How much force is exerted by a magnetic field on a stationary electric dipole ?
Ans, Zero
Q 6. Explain, how moving charge is a source of magnetic field.
Ans. A moving charge produces an electric current which in turn causes the magnetic field.
Q 7. How does a current loop behave like a magnetic dipole ?
45
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Ans. One face of current loop behaves as a south pole and the other face as north pole. Therefore, the loop behaves as a magnetic
dipole.
Q 8. An electron beam projected along + X-axis, experiences a force due to a magnetic field along the + Y-axis. What is
the direction of the magnetc field ?
Sol. The direction of magnetic field is along the + Z-axis, as per right hand rule or Fleming‘s Left hand rule.
Q 9. An electron and a proton, moving parallel to each other in the same direction with equal momenta, enter into a
uniform magnetic field which is at right angles to their velocities. Trace their trajectories in the magnetic field.
Ans. Both electron and proton will trace circles of equal radii but in opposite senses.
Q 10. Under what condition does an electron moving through a magnetic field experiences maximum force ?
Ans. F = q v B sin θ ; F is maximum if sin θ = 1 or = 90°. So, if the electron is moving perpendicular to the direction of the
magnetic field, it experiences a maximum force.
Q 11. Under what condition is the force acting on a charge moving through a uniform magnetic field minimum ?
Ans. F = q v B sin θ ,F is minimum if sin θ =0 or = 0° or 180°. So, if the charged particle is moving parallel or antiparallel to the
direction of magnetic field, it experiences minimum force.
Q12.




 
The force F experienced by a particle of charge 'q' moving with a velocity v in a magnetic field B is given by F  q  vB  .
Which pairs out of these vectors are always at right angles to each other?



Ans. F is perpendicular to v and B
Q 13. In what ways electric and magnetic fields are different ?
Ans. (i) Electric field is due to charges at rest as well as in motion, whereas magnetic field is due to a magnet or current through a
conductor.
(ii) The strength of electric field at a point decreases with the dielectric medium but the strength of magnetic field increases when
a permeable medium is inserted there.
(iii) The electric lines of force representing the electric field do not form a closed path whereas the magnetic lines of force form a
closed path.
Q 14. An electron is passing through a field but no force is acting on it. Under what conditions is it possible, if the motion
of the electron be in the (i) electric field (ii) magnetic field ?
Sol. In electric field, there is always a force on the moving electron opposite to the direction of field. Thus the force will be zero
only if field is zero. (ii) In magnetic field, the force acting on a moving electron is
F = q v B sinθ , it is zero if θ = 0° or 180°.
i.e. the electron is moving parallel to the direction of magnetic field.
Q 15. The energy of a charged particle moving in a uniform magnetic field does not change. Explain.
Sol. When a charged particle is moving in a uniform magnetic field, it experiences a force in a direction, perpendicular to its
direction of motion and the speed of the charged particle remains unchanged and hence its kinetic energy remains same.
Q 16. In what respect the electric force and magnetic force on a charged particle differ when the charged particle is
subjected to (i) electric field and (ii) magnetic field.
Ans. 1. The electric force on the charged particle is collinear with the electric field where as magnetic force is always
perpendicular to the magnetic field.
2. The electric force on the charged particle in electric field is independent of the state of rest or of motion of the charged particle
in the electric field, whereas magnetic force only acts when charged particle is moving in the magnetic field, as magnetic force is
velocity dependent.
3. The electric force accelerates the positively charged particle along the direction of electric field. The magnetic force
accelerates the moving charged particle perpendicular to the direction of magnetic field.
4. The electric force changes the speed of the charged particle and hence changes its K.E., whereas the magnetic force does not
change the speed of the charged particle and hence no change in its K.E.
Q 17. State properties of the material of the wire used for suspension of the coil in a moving coil galvanometer.
Ans.. The properties of the material of the wire used for suspension of the coil in a moving coil galvanometer are as follows :
1. It should have low torsional constant i.e. restoring torque per unit twist should be small.
2. It should have high tensile strength.
3. It should be a non-magnetic substance.
46
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
4. It should have a low temperature coefficient of resistance.
Q 18. Can you accelerate neutrons by cyclotron ? Explain.
Sol. Neutrons being neutral in character can not be accelerated by cyclotron because a cyclotron can accelerate only charged
particles. Cyclotron is most suitable to accelerate positively charged particle like protons, alpha particles etc.
Q 19. Why should a solenoid tend to contract when a current passes through it ?
Sol. We know that two parallel conductors carrying currents, in the same direction attract each other and in opposite directions
repel each other. Therefore, when current is passed through the coil of a solenoid, the parallel currents in the various turns of
solenoid flow in the same directions. As a result of it, the various turns start attracting one another and solenoid tends to contract.
Q 20. What is a radial magnetic field ? How it has been achieved in moving coil galvano- meter ?
Sol. Radial magnetic field is that field, in which the plane of the coil always lies in the direction of the magnetic field. A radial
magnetic field has been achieved by (i) properly cutting the magnetic pole pieces in the shape of concave faces, (ii) using a soft
iron core within the coil.
Q 21. Why is phosphor bronze alloy preferred for the suspension wire of a moving coil galvanometer ?
Sol. The suspension wire of phosphor bronze alloy is preferred in moving coil galvanometer because it has several advantages :
(i) Its restoring torque per unit twist is small. Due to it, the galvanometer is very sensitive.
(ii) It has great tensile strength so that even if it is thin, it will not break under the weight of the coil suspended from its end.
(iii) It is rust resisting. Hence it remains unaffected by the weather conditions of air in which it is suspended.
Q 22. A current carrying loop, free to turn, is placed in a uniform magnetic field. What will be its orientation, relative to
B, in the equilibrium state ?
Sol. A current carrying loop behaves as a magnetic dipole of magnetic moment M, acting perpendicular to its plane. The torque
on the current loop of magnetic dipole moment M in the magnetic field B is = MB sinθ .
where is the angle between M and S. The system is in stable equilibrium if torque is zero, which is so if θ= 0°. This is possible if
B is perpendicular to the plane of the coil.
Q 23. Can we increase or decrease the range of a given ammeter ?
Sol. We can increase the range of the given ammeter by using a suitable shunt in parallel with the ammeter but we can not
decrease the range of the ammeter .
Q24. Can we decrease or increase the range of the given voltmeter ?
Sol. We can increase the range of the given voltmeter by putting a suitable high resistance in series with the voltmeter.
We can decrease the range of the given voltmeter by putting a suitable resistance in parallel with the high resistance already in
series with the galvanometer working as voltmeter, so that the effective resistance of the voltmeter multiplied with the current (I )
gives the required potential difference to be measured.
Q 25. Why ammeter is connected in series and voltmeter in parallel in the circuit ?
Sol. An ammeter is a low resistance galvanometer. It is used to measure the current in amperes. To measure the current of a
circuit, the ammeter is connected in series in the circuit so that the current to be measured must pass through it. Since, the
resistance of ammeter is low, so its inclusion in series in the circuit does not change the resistance and hence the main current in
the circuit.
Voltmeter: A voltmeter is a high resistance galvanometer. It is used to measure potential difference between two points of the
circuit in volts. To measure the potential difference between the two points of a circuit, the voltmeter is connected in parallel in
the circuit. The voltmeter resistance being high, it draws minimum current from the main circuit and the potential difference to be
measured is not affected materially,
Q26. Which one has lowest resistance; ammeter, voltmeter and galvano- meter ? Explain.
Sol. Ammeter has the lowest resistance. A galvanometer is converted into an ammeter by using a suitable low resistance shunt in
parallel of the galvanometer. The effective resistance of the net work when resistors are connected in parallel becomes less than
the individual resistance. Hence resistance of ammeter is less than that of galvanometer.
Q27. What is the path of a charged particle moving in a uniform magnetic field with initial velocity
(i) parallel to the field ? (ii) perpendicular to the field? (iii) at some angle with the direction of magnetic field.
Ans. (i) A straight line, (ii) a circular in a plane perpendicular to the field, (iii) a helical path with axis parallel to direction of
magnetic field.
Q 28. A proton is moving with a velocity in a direction of the magnetic field B. Write down the magnitude of the force
acting on the proton.
Ans. Zero.
Q 29. What is meant by cyclotron frequency ?
47
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Ans: The cyclotron frequency is given by
Bq . This is independent of the radius of the circular path and velocity of the
2 m
charged particle in the two dees of cyclotron.
Q30. What type of force is acting between two parallel wires carrying current in the same direction ? What happens if
one of the currents is reversed ?
Ans. Force of attraction. When one current is reversed, the force acting between two parallel wires carrying currents becomes
repulsive.
Q. 31. What is the magnitude of torque which acts on a coil carrying current placed in a uniform radial magnetic field ?
Ans. Torque, τ= nBIA, where the terms have their usual meanings.
Q. 32. What is a dead beat galvanometer ?
Ans. Dead beat galvanometer is one in which the coil comes to rest at once after the passage of current through it. The deflection
can be noted in no time.
Q. 33. Why is the coil wrapped on a conducting frame in a galvanometer ?
Ans. Eddy currents in conducting frame help in stopping the coil soon i.e. in making the galvanometer dead beat.
Q. 34. What is the function of soft iron cylinder between the poles of a galvanometer ?
Ans. It concentrates the magnetic field and helps in making the magnetic field radial.
Q.35. What is the function of the radial magnetic field in the moving coil galvanometer ?
Ans. It helps the arm of the couple and hence the torque on coil always the same in all positions. This arrangement provides
linear scale to the galvanometer.
Q. 36. Why are the conductors used in cyclotron are called Dees ?
Ans, It is due to their D-shape.
Q37. Why are pole pieces of galvanometer made concave ?
Ans. To have a uniform, strong and radial magnetic field.
Q38. What is the nature of magnetic field in moving coil galvanometer ?
Ans. A radial magnetic field.
Q 39. What is the resistance of ideal ammeter ?
Ans. Zero.
Q 40. What is the resistance of ideal voltmeter ?
Ans. Infinite.
Q 41. What is meant by figure of merit of a galvanometer ?
Ans. Figure of merit of galvanometer is defined as the amount of current which produces one scale deflection in the
galvanometer.
Q.42. What is the effective resistance of ammeter if a shunt of resistance S
galvanometer of resistance G ?
Ans. GS /(G + S).
is used across the terminals of the
Q43. Which has greater resistance (a) milliammeter or ammeter ? (b) milli voltmeter or voltmeter?
Ans. The resistance of milliammeter is greater than ammeter. The resistance of voltmeter is greater than millivoltmeter.
Q 44. If the distance between two parallel current carrying wires is doubled, what is the force between them ?
Ans. The force acting on one wire due to currents through two wires is inversely proportional to the distance
between them. Thus the force becomes 1/2 times if the distance between the wires is doubled.
Q 45. On what interaction the principle of galvanometer is based ?
Ans. The principle of galvanometer is based on the interaction of current and magnetic field.
Q 46. By mistake a voltmeter is connected in series and an ammeter is connected in parallel, with a resistance in an
electrical circuit What will happen to the instruments ?
48
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Ans. Voltmeter resistance being very high, when used in series of the circuit, the effective resistance of the circuit becomes very
high. Due to which the current in the circuit becomes very low. As voltmeter measures potential difference between the two
points, it will show the reading but will not be damaged. The ammeter resistance being very low and connected in parallel of the
circuit, the weak current flowing through the circuit will pass through the ammeter. Due to which the ammeter will not show
appreciable deflection. Thus ammeter also will not be damaged.
Q 47. What is the ratio of electric and magnetic forces between two moving charges ?
Ans. Fe/Fm= 1027
Q 48. What is the basic principle of working of cyclotron ?
Ans. The working of the cyclotron is based on the fact that a heavy positively charged particle can be accelerated to a sufficiently
high energy with the help of smaller values of oscillating electric field, by making it to cross the same electric field time and
again with the use of strong magnetic field.
Q 49. The pole of a magnet is brought near a stationary charge, will the pole experience any force ?
Ans. No, the stationary charge does not produce any force.
Q 50. An ammeter and a milliammeter are converted from the same galvanometer. Out of the two, which current
measuring instrument has smaller resistance ?
Ans. Ammeter has lower resistance than milliammeter as the shunt to be used across galvanometer is of lower resistance for
ammeter than that of milliammeter.
Q.51. Which loop will experience greater torque ? Give reasons.
Ans. For a wire of given length, the circular loop encloses greater area than the square loop. As, torque   IAB
so the circular current loop will experience greater torque in the magnetic field.
ie.   A
Q 52. Can moving coil galvanometer be used to detect an a.c in a circuit ? Give reason.
Ans. A moving coil galvanometer cannot be used to detect a.c. in a circuit, since it measures the average value of current and the
average value of a.c. over a complete cycle is zero.
Q 53. What is shunt ? State its S.I. Units.
Ans. A shunt is a low resistance which when connected in parallel to the galvanometer, protects it from the strong current. The
S.I. unit of shunt is ohm.
Q 54. What information would you wish to have about the galvanometer before you convert the galvanometer
into an ammeter or voltmeter ?
Ans. Before converting the galvanometer into an ammeter or voltmeter we require the following two informations.
(i) The resistance of galvanometer and (ii) the current for full scale deflection of galvanometer.
Q 55. Of the two identical galvanometers one is to be converted into an ammeter and another into a milli ammeter.
Which of the shunts will be of larger resistance ?
Ans. The shunt of milliammeter should have larger resistance. It is so because the current measuring range of miliiammeter is
less than that of ammeter. Therefore the larger portion of the main current must pass through galvanometer coil and less portion
of current through shunt fo a converted
galvanometer into ammeter/ miliiammeter. It is only possible if the resistance of shunt used is of larger value in miliiammeter
than of ammeter.
Q 56. Why do we not use galvanometer as an ammeter ?
Ans. A galvanometer shows a full scale deflection with a very small current. Hence a galvanometer can measure limited current.
Therefore as such, a galvanometer can not be used as an ammeter, which can measure the given current.
Q 57. Why should an ammeter have a low resistance and a high current carrying capacity ?
Ans. An ammeter is used to measure the current. It can measure the current of the circuit if connected in series of the circuit. The
ammeter connected in series of the circuit can measure the current and will not disturb the current of the circuit if its resistance is
low and current carrying capacity is high.
Q 58. Why should a voltmeter have a high resistance and a low current carrying capacity ?
Ans. Voltmeter is a high resistance galvanometer. It is used to measure potential difference between two points of the circuit. To
measure a potential difference between two points of a circuit, the voltmeter is connected in parallel to the circuit across those
49
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
two points. The potential differences between those two points will not be affected if there is no change in the current flowing
through the circuit between those two points, which will be so if practically no current flows through voltmeter. The same is
possible if voltmeter has a high resistance and a low current carrying capacity. That is why a voltmeter should have high
resistance and low current capacity.
Q 59. Out of an ammeter and a voltmeter, which of the two has higher resistance and why ?
Ans. Ammeter is obtained when a suitable shunt is connected in parallel to the galvanometer. Since the effective resistance in
parallel becomes less than the individual resistance. The resistance of shunt is also low, therefore the ammeter resistance is low.
Voltmeter is obtained when a suitable resistance is connected in series to the galvanometer. Since the effective resistance in series
becomes more than the individual resistance, hence the voltmeter resistance is high.
Thus the voltmeter resistance is higher than that of ammeter.
.Q 60. Suppose a helical spring is suspended from the roof of a room and very small weight is attached to its lower end.
What will happen to the spring when current is passed through it Give reason to support your answer.
Ans. The spring will contract. In the two adjoining windings of the spring the current is in the same direction, hence there will be
attraction between them. Due to which the weight attached at the end of spring is lifted upwards.
Q 61. How does (i) an ammeter (ii) voltmeter differ from a galvanometer ?
Ans. (i) Ammeter is a low resistance galvanometer. When a low resistance shunt is connected across galvanometer, it becomes
ammeter.
(ii) Voltmeter is a high resistance galvanometer. When a suitable resistance is connected in series of galvanometer, it becomes
voltmeter.
Q 62. What is an ammeter ? How is it used in an electric circuit ? How does it differ from a voltmeter.
Ans. Ammeter is a low resistance galvanometer. It is connected in series of the circuit. The resistance of ammeter is low and of
voltmeter is high. Ammeter is connected in series and voltmeter in parallel to the circuit.
Q 63. What are the limitations of cyclotron?
Ans: (i) Cyclotron is suitable only for accelerating heavy particles such as protons, particles, etc. It is not suitable for accelerating
electrons.
(ii) Cyclotron cannot accelerate uncharged particles (e.g., neutrons).
(iii) For very high kinetic energy (e.g. 500 GeV), it is impossible to design magnetic field system.
Q 64. Define one Ampere of current.
Ans. One ampere is defined as that current flow ing in each of the two long parallel conductors lm apart, which results in a force
of exactly 2 x 109 N per metre length of each conductor.
Q 65. What is current sensitivity of a moving coil galvanometer ? Write its expression. On what factors does the current
sensitivity of the galvanometer depend upon.
Ans. Current Sensitivity: If current I is passed through a galvanometer and deflection produced is 0, then current sensitivity of
the galvanometer is
 i.e. current sensitivity is the deflection produced in the galvanometer when a unit current is passed
I
through it.
It means that in order to have high current sensitivity n, B and A must be large while k should be small.
(i) n must be large: There is, however, an upper limit to n. If n is made very large, resistance of the galvanometer increases
sufficiently as well as the coil becomes bulky. This tends to decrease the sensitivity.
(ii) B must be large: This is achieved by using a narrow air gap and a strong permanent magnet (typically B = 0.5 T). The
additional advantage of high value of B is that the instrument is unaffected by external magnetic fields (e.g. Earth‘s field = 4 x
10-5 T).
(iii) Area A must be large: There is an upper limit to it because it must not be so large that the instrument becomes bulky.
Further, a coil of large area swings about its equilibrium position for a long time.
Q 66. Explain how will you convert a galvanometer into an Ammeter.
Ans:To convert a galvanometer into an Ammeter a resistance of low value is connected in parallel with
the galvanometer.
Q 67. How does the intensity of magnetisation of a paramagnetic material vary with increasing applied magnetic field?
Ans. Intensity of magnetization of a paramagnetic material is directly proportional to the applied magnetic field.
Q 68. Distinguish between a magnetic dipole and an electric dipole.
Magnetic dipole
Electric dipole
50
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
1.
It consists of two equal and opposite poles which
do not have any separate existence.
1.
It consists of two equal and opposite charges which
have a separate existence.
2.
Poles of a magnetic dipole do not move.
2.
Charges of an electric dipole can move.
Q69. What are the limits of earth‟s magnetic field ?
Ans. Earth‘s magnetic field extends up to a distance of 32000 km, which is about five times the radius of earth.
Q70. A magnetic dipole is situated in the direction of a magnetic field. What is its potential energy ? If it is rotated by
1800, then what amount of work will he done ?
Ans:. P.E. of dipole is zero. Work done , MB (cos 0° - cos 180°) = MB (1 + 1) = 2 MB.
Q 71 . A rectangular current loop is in a magnetic field in any orientation. Will any work he done in rotating the loop
about an axis perpendicular to its plane ?
Ans. No. This is because forces act perpendicular to the arms of the loop and not parallel to them. Hence there is no couple of
forces acting on the loop, which would rotate the loop about an axis perpendicular to its plane.
Q 72. What is a magnet ?
Ans. A magnet is a material which has attractive and directive properties.
Q73. what is a natural magnet ?
Ans. A natural magnet is a naturally occurring ore of iron, which has attractive and directive properties.
composition is Fe3O4.
Its chemical
Q74. Name some magnetic and non-magnetic substances.
Ans. Iron, cobalt, nickel, steel etc. are magnetic substances. Brass, paper, wood etc. are non- magnetic substances.
Q 75. What is a magnetic dipole ?
Ans. Two unlike poles of equal strength separated by some distance form a magnetic dipole.
Q 76. Where is the vertical component of earth‟s magnetic field zero ?
Ans. At earth‘s magnetic equator.
Q.77. Can we have a magnet with a single pole ?
Ans. No, because unlike poles of equal strength exist together.
Q. 78. Are the two poles of a magnet equally strong ?
Ans. Yes, always.
Q 79. What are the units of dipole moment in S.I. system ?
Ans. Ampere-metre..
Q 80. Is dipole moment a scalar or vector ?
Ans. It is a vector.
Q81. What is the direction of dipole moment ?
Ans. From south pole to north pole of the dipole.
Q. 82. What is the S.I, unit of magnetic field strength ?
Ans. Tesla .
Q. 83. Magnetic lines of force are endless. Comment.
Ans. This is because magnetic lines of force are continuous closed loops.
Q. 84. What is the angle of dip at a place where horizontal and vertical components of
earth‟s field are equal ?
Ans. 45°
51
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Q 85. A magnetic needle free to rotate in a vertical plane, orients itself with its axis vertical at a certain place on the earth.
What are the values of (a) Horizontal component of earth‟s field? (b) angle of dip at this place.
Ans. H = 0 and θ = 90°.
Q 86. Give two examples of magnetic dipole.
Ans. Every atom of para and ferro magnetic substances; a loop of current.
Q87. Write mathematical form of tangent law in magnetism.
Ans. F = H tanθ, where F and H are strengths of two magnetic fields acting perpendicular to each other and θ is the angle which
freely suspended magnet makes with the direction of H.
Q 88. Can ever there be a magnet (a) with no pole (b) with two similar poles (c) with three poles ?
Ans. (a) Yes. there can be a magnet with no pole e.g. in case of a toroid.
(b) Yes, there can be a magnet with two similar poles when like poles of two magnets are glued together.
(c) Yes, there can be a magnet with three poles.
Q 89. If a compass box and a dip circle were to be taken to the magnetic north pole of earth, what would one observe with
regard to directions of their respective needles and why ?
Ans. The needle of compass box shall not necessarily stand along north south direction. It may point along any arbitrary
direction.
The needle of a dip circle shall stand vertical with north pole pointing upwards.
Q 90. What is a magnetic field line?
Ans. The path along which the compass needles are aligned is known as magnetic line of force.
Following are some of the important properties of magnetic lines of force:
1. Magnetic lines of force are closed continuous curves, we may imagine
them to be extending through the body of the magnet.
2. Outside the body of the magnet, the direction of magnetic lines of force, is from
north pole to south pole.
3. The tangent to magnetic line of force at any point gives the direction of magnetic field strength at that point.
4. No two magnetic lines of force can intersect each other.
5. Magnetic lines of force contract longitudinally and they dilate laterally.
6. Crowding of magnetic lines of force represents stronger magnetic field and vice-versa.
52
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
(2) when =900 , i.e, uniform magnetic field is along
the surface then
 = BA cos 900  0
( cos 900  0)
Thus, the magnetic flux through a given surface
is zero when  =900
4.16. Faraday‟s laws of electromagnetic induction
First Law:
When the magnetic flux linking a conductor or coil
changes, an e.m.f. is induced in it.
Second Law: The induced emf lasts as long as there is
relative motion between the magnet and the conductor.
UNIT IV. Electromagnetic
Induction and Alternating
Currents
Electromagnetic induction is the phenomenon of
production of electric field with the help of varying
magnetic field.
Magnetic flux: Magnetic flux over a given surface is
defined as the number of magnetic lines of force crossing
through that surface.
 
  B. A  BA cos
Third Law:
The magnitude of the e.m.f. induced in a conductor or coil
is directly proportional to the rate of change of flux
linkages.
e  N
d
dt
4.17. Lenz‟s law.
Lenz‟s law: Emil Lenz, a German scientist gave the
following simple rule (known as Lenz‘s law) to find the
direction of the induced current:
The induced current will flow in such a direction so as to
oppose the cause that produces it.
4.18. EDDY CURRENTS
When a conductor moves in a magnetic field or
a changing magnetic field, the induced currents
circulate throughout the volume of the metal.
Because of their general circulating nature, these are
called eddy currents.
Undesirable Effects: Eddy currents induced in the
armature of a motor or generator produce heat. This results
in the waste of energy and the rise of temperature of the
machine. To reduce the eddy currents, the armature is split
into thin sheets (called laminations) in planes parallel to the
magnetic field.
Conditions for maximum and minimum magnetic flux:
(1) when  =00 , i.e, uniform magnetic field is acting
Useful applications:
perpendicular to the plane of the surface, then
(i) Eddy current damping:
They can reduce the oscillations of a vibrating system.
 =BA cos 0
or,  =BA=maximum
(  cos 0  1)
Thus, magnetic flux through a given surface is
0
maximum i.e, maximum number of
magnetic lines of force passes through the given
surface when  =00
(ii) Induction heating: The heating effect of eddy currents
can be used to heat/melt those substances which are
conductors of electricity.
(iii) Electro-magnetic Brakes: Eddy current braking can
be used to control the speed of electric trains. In order to
reduce the speed of the train, an electromagnet is turned on
53
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
that applies its field to the wheels. Large eddy currents are
set up which produce the retarding effect.
Units of Self Inductance
SI unit of self inductance is henry (H).
Q. What is self induction?
since,
L

dI
dt
Sol. When current in a coil increases or decreases, there is a
change in magnetic flux linking the coil. Hence, an e.m.f. is
induced in the coil. This is called self-induced e.m.f. and
the process is called self-induction.
Also, L 
Q. Define self inductance.
i.e, 1H  1 VA-1s  1Wb A1
1volt
 1VA-1s
1 ampere second -1
1 weber
1henry (H)=
 1Wb A1
1 ampere
1henry (H)=

I
Smaller units of Self inductance are : 1mH=10-3 H
and 1 H  106 H .
Sol. The property of a coil (or a circuit) by virtue of which
it opposes any change in the amount of current flowing
through it is called its self-inductance.
4.19. Coefficient of Self Induction or Self Inductance
Let I be the current flowing through a coil then
the magnetic flux ( ) linked with the coil is found
to be proportional to the strength of the current
(I)
i.e,   I or ,   L I       (1)
where L is the constant of proportionality
and is known as Co-efficient of self induction
or self inductance .
If I = 1, then from equation (1)
L= 
Thus, Co-efficient of self induction of a
coil is defined as the magnetic flux linked
with a coil when unit current flows through
it.
4.20. Mutual induction
It is the phenomenon of inducing e.m.f. in a coil due to
increase or decrease of current (rate of change of current)
in a nearby coil.
s  I p or s =M I p (I p =current in primary coil)
here, M=constant known as co-efficient of mutual
induction or mutual inductance.
If I p =1, then M= s
Thus, mutual inductance of two coils or circuits can be
defined as the magnetic flux linked with the secondary coil
due to the flow of current in the primary coil.
Unit of mutual inductance is 1H=1WbA-1=1V A-1s
5.0. To verify that Lenz‘s law is in accordance with
the law of conservation of energy.
Also, according to Faraday's Law of
electromagnetic induction, induced
e.m.f in the coil is given by
d
.
dt
using equation (1), we get
d
dI
 =  ( L I ) or ,  =  L
   (2)
dt
dt
dI
If 
 1 i.e, rate of decrease of current
dt
is unity, then from equation (2), we get
L
 = 
Thus, Co-efficient of self induction of a coil
is defined as the induced e.m.f produced in the
coil through which the rate of decrease of current
is unity.
According to Lenz‘s law, the direction of the induced
current will be such so as to oppose the cause that produces
it.Lenz‘s law directly follows from the law of conservation
of energy i.e. in order to set up induced current, some
energy must be expended. For example, when the N-pole
of the magnet is approaching the coil, the induced current
will flow in the coil in such a direction that the left-hand
face of the coil becomes N-pole. The result is that the
motion of the magnet is opposed. The mechanical energy
spent in overcoming this opposition is converted into
electrical energy which appears in the coil. Thus Lenz‘s
law is consistent with the law of conservation of energy.
54
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Alternating current
The current that changes magnitude with time and polarity
reverses periodically is known as alternating current (A.C).
The sinusoidal alternating current is represented by I=I0sin
 t, where I0=peak value of a.c. and angular frequency
2

 2 .
T
The sinusoidal e.m.f. is given by E  E0 sin t
Advantages and Disadvantages of AC
1.
2.
3.
4.
5.
A.C. is economical in transmission.
A.C. can be easily converted into D.C.
A.C. can be transmitted to distant places without
loss in electric power.
A.C. voltage can be stepped up or stepped down.
A.C. machines are stronger than D.C.
Q. What is the importance of r.m.s current?
Ans. . An alternating voltage or current is always specified
in terms of r.m.s.
values. Thus an alternating current of 10 A is one which
has the same
heating effect as 10 A d.c. under similar conditions.
(i)
The domestic a.c. supply is 230 V, 50 Hz. It
is the r.m.s. or effective value. It means that
the alternating voltage available has the
same heating effect as 230V d.c. under
similar conditions. The equation of this
alternating voltage is:
E  E0 sin t
 230  2 sin 2  50  t  E0 
 230  2 sin 314t
(ii)
(iii)
1.
2.
3.
4.
A.C can be more dangerous than D.C in terms of
its attractive nature and also because its
maximum value is √2 times its effective value.
A.C cannot be used in the electrolytic processes
such as electroplating, electrotyping and electrorefining etc. where only D.C is used.
A.C in a wire is not uniformly distributed
throughout its cross-section. The a.c density is
much greater near the surface of the wire than
that inside the wire. This concentration of a.c
near the surface of the wire is called the skin
effect.
Markings on the scales of A.C meters are not
equidistant for small measurements.
2 Ev 
5.2.
When we say that alternating current in a
circuit is 5 A, we are specifying the r.m.s.
value. It means that the alternating current
flowing in the circuit has the same heating
effect as 5 A does under similar conditions.
A.C. ammeters and voltmeters record r.m.s.
values of current and voltage respectively.
Alternating e.m.f Applied to a
Resistor
5.1. Root Mean Square ( R.M.S) Value of
Alternating Current
Root mean square value of a.c is defined as that steady
current which produces the same amount of heat in a
conductor in a certain time interval as is produced by the
a.c in the same conductor during the time period T(i.e. full
cycle).
It is represented by Irms
or,
I rms 
Io
2
 0.707 I o
Q. Define root mean square value of a.c.
Ans. It is that steady current which produces the same
amount of heat in a
conductor in a certain time interval as done by the a.c. in
the same conductor during the time T(i.e. full cycle).
55
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
An A.C source connected to a resistor of
As there is no potential drop across the circuit
resistance R. Such a circuit is known as a resistive
so,
circuit. The applied alternating e.m.f is given by
E=E 0 sin t
      (1)
dI 
dI

E+   L   0 or , L  E
dt 
dt

dI E
or , 
dt L
Using equation (1), we get
Let I be the current in the circuit at any instant t.
The potential difference across the resistor =I R.
So,
E=I R
E
or,
I= , using equation (1), we get
R
E sin t
I= 0
or , I  I 0 sin t     (2)
R
E
where I 0  0 is the peak value of alternating current.
R
Comparison of equation (1) and (2) shows that the
current and e.m.f across the resistor are in phase.
5.3.Alternating e.m.f Applied to an
Inductor
E
dI E 0

sin t or , dI  0 sin t dt
dt
L
L
    (2)
Integrating both sides , we get
E
E
 dI   L0 sin t dt  L0  sin t dt
E  cos t  E 0
or , I  0  

  cos t 
L 
  L


Since,   cos t   sin  t   and peak value
2

E0
I0 
L
 I  I 0 sin t   / 2 
Thus current lags behind emf by an angle of π/2
Inductive Reactance (XL)
Eo
E
with I0  o , we conclude
L
R
that, (L ) has the dimensions of resistance.
The term (L ) is known as inductive reactance
represented by X L .
Comparing I0 
An alternating e.m.f applied to an ideal inductor
of inductance L.Such a circuit is known as purely
inductive circuit.
The alternating e.m.f applied is given by
E=E 0 sin t
which opposes the growth of current in the
circuit.
Behaviour of pure inductor in case of d.c and a.c
Inductive reactance,X L  L  L  2 v
     (1)
The induced e.m.f across the inductor =  L
The inductive reactance is the effective opposition
offered by the inductor to the flow of current in the
circuit.
In S.I, unit of capacitive reactance is ohm.
dI
dt
For , d .c
v0

XL  0
Clearly, inductor offers no opposition to the flow
of d.c. Hence d.c can flow easily through an inductor.
For, a.c v = finite

X L  finite value
Thus, inductor offers finite opposition to the
flow of a.c.
56
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
5.4. Alternating e.m.f. Applied to a Capacitor
5.5. L,C and R series circuit with alternating
supply
Alternating e.m.f applied to a capacitor. Such a
A circuit containing inductor of pure inductance
( L), capacitor of pure capacitance (C) and
resistor of resistance (R), all joined in series. Let
E be the r.m.s value of the applied alternating
e.m.f to the LCR circuit.
circuit is known as purely capacitive circuit.
The alternating e.m.f applied across the capacitor
is given by
E=E 0 sin t
Let q be the charge on the capacitor at any instant.
 Then potential difference across the capacitor,
q
c
Vc 
q
 E  E0 sint.
c
q  E0c sint.
But , Vc  E or

Now,

I
E0
dq d
  E0c sint   E0c (cos t )
dt dt
1/ c 
cos t.


sin ce, cos t  sin  t  
2



 I  I o sin  t  
2

Hence in a electric circuit containing C only, the current is
Let I be the r.m.s value of current flowing through
all the circuit elements.
The potential difference across inductor,
VL  IX L       (1)
ahead of EMF by an angle of 900 .
(leads current I by an angle of  /2)
The potential difference across C, VC  IX C
Capacitive Reactance (Xc)
       (2)
(lags behind the current I by an angle of  /2)
The potential difference across R, VR  IR
Comparing I o 
E0
E
with I o  0 , we conclude
R
 1 


 C 
 1 
that 
 has the dimension of resistance.
 C 
 1 
The term 
 is known as Capacitive reactance
 C 
(X C ).
The capacitance reactance is the effective
opposition offered by a capacitor to the flow of
current in the circuit.
In S.I unit of inductive reactance is ohm.
Behaviour of a capacitor in case of d.c. and
a.c.
       (3)
( in phase with the current)
The resultant of VR and VL  VC  is given by OH .
OH 2  OA2  OD 2
 OH  OA2  OD 2 
E
2
I 2 R 2   IX L  IX C   I R 2   X L  X C 
2
2
E
2
  R2   X L  X C 
I
E
But
Z
I
Im pedence
I 
1
1
Capacitive reactance, X C 

C C  2 v
For , D.C., v  o
1

X C   .
0
Thus, capacitor offers infinite resistance to the
flow of d.c. So, d.c cannot pass through a capacitor,
however small the capacitance of the capacitor is,
For A.C., v  finite
1

XC 
 smaller value.
finite value
VR 2  VL  VC 
E

Z
Z
R2   X L  X C 
E
R2   X L  X C 
2

2
E
1 

R  L 

C 

2
2
1 

 X L   L and X C  C 
The current (I) in a series LCR circuit is given by
I
E
E


2
Z
R2   X L  X C 
E
1 

R2    L 

C 

2
57
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Impedence and Admittance
Working:
The effective opposition offered by LCR circuit to
alternating current is known as Impedence.
Admittance (A) = . The reciprocal of impedence is
called admittance.
Impedence triangle
Impedence (Z) of LCR circuit can be represented
diagrammatically by impedence triangle as shown in Fig
1 
2

Im pedence Z  R 2   X L  X C   R 2    L 
C
 

2
5.6. LC oscillator
Electrical oscillations produced by the exchange of energy
between a capacitor which stores energy and an inductor
which stores magnetic energy are called LC oscillations. A
circuit containing inductor and capacitor is called a tank
circuit or LC oscillator.
When the capacitor is discharged completely, the energy is
stored in the magnetic field around
the inductor I. In other words, electric energy is
completely converted into magnetic energy.
When the magnetic field energy becomes maximum, the
capacitor begins to re-charge itself in
the opposite direction. Now the energy stored in the
magnetic field is converted into the energy stored in the
electric field of the capacitor.
When the capacitor is fully re-charged, whole of the energy
is stored in the electric field of the capacitor. At this instant,
the capacitor is again discharged through the inductor. The
current flows through the inductor and energy is stored in
the magnetic field around it .
Now again the capacitor is re-charged in the opposite
direction and the energy is stored in the electric field of the
capacitor.
In the whole process, there is exchange of electric energy
and magnetic energy. This exchange of energy from electric
form to magnetic form gives rise to oscillations called LC
oscillations.
Frequency of LC oscillator will be  
1
2 LC
POWER PACKED QUESTIONS
Q1. What type of oscillations is produced by a LC
oscillator? Why does a LC oscillator become warmer?
Ans. The oscillations produced by LC oscillator is a
damped oscillation. In such oscillation the magnitude of
oscillations decreases with increase in time. It is because
the resistance of the circuit increases with the increase in
temperature since the LC circuit becomes warmer.
The LC circuit becomes warmer due to the fact that energy
is dissipated in the form of heat.
Q 2. What is electrical resonance for LCR series
circuit? Find the expression for the resonant frequency
of LCR circuit. What is the use of such circuit?
Ans:Electrical resonance is said to take place in a series
LCR circuit when the circuit allows maximum current for a
given frequency of the source of alternating supply for
which capacitive reactance becomes equal to the inductive
reactance.
Q 3. Will an induced current develop in a conductor
moved in a direction parallel to magnetic field ?
58
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Ans. No. This is because when the conductor is moved in a
direction parallel to magnetic field, amount magnetic flux
linked with the conductor does not change.
Q4. A vertical metallic pole falls down through the
plane
of
the
magnetic
meridian. Will any e.m.f. be induced between its ends ?
Ans. No e.m.f. will be induced across the ends of the pole,
as it intercepts neither
the horizontal component H nor the vertical component V
of earth‘s field.
Q5. A coil is removed from a magnetic field (i), rapidly
(ii) slowly. In which case, more work will be done ?
Ans. Work done will be more when the coil is removed
rapidly. This is because in that
case opposing e.m.f. induced in the coil will be more.
Q6. A copper ring is suspended by a thread in a
vertical plane. One end of a magnet is brought
horizontally towards the ring. How will the position of
ring be affected ?
Ans. The ring will move away from the magnet. This is
because as per Lenz‘s law, an e.m.f. is induced in the ring
with N pole at the face of the ring towards the magnet.
Force of repulsion is responsible for motion of the ring
away from the magnet.
Q 7. A bar magnet falls through a metal ring. Will its
acceleration be equal to „g‟ ?
Ans. No, acceleration of the magnet will not be equal to g.
It will be less than „g‟
This is because as the magnet falls, amount of magnetic
flux linked with the ring changes. An induced current is
developed in the ring which opposes the downward motion
of the magnet. After the magnet has crossed the metal ring,
amount of magnetic flux linked with the ring goes on
decreasing. An induced current developes in the ring and
opposes the fall of the magnet. Therefore, downward
acceleration of the magnet continues to be less than „g‟
Q 8. A lamp in a circuit consisting of a coil of large
number of turns and a battery does not light up to full
brilliance instantly on switching on the circuit. Why ?
Ans. When the circuit is switched on, current increases
through the lamp and also through the coil. An induced
e.m.f. developes in the coil which opposes the growth of
current. As the current takes time to grow to maximum
value, the lamp does not light instantly upto full brilliance.
Q 9. As soon as current is switched on in a high-voltage
wire, the bird sitting on it flies away. Why ?
Ans. When current is switched on, induced current flows in
the body of the bird. Its wings experience mutual repulsion
on account of opposite currents in them. Therefore, the bird
flies.
Q 10. A ring is fixed to the wall of a room. When south
pole of a magnet is brought near the ring. What shall be
the direction of induced current in the ring ?
Ans. The induced current is clockwise as seen from the side
of the magnet.
Q 11. Does change in magnetic flux induce e.m.f. or
current ?
Ans. emf is always induced, but current will be induced
only when the circuit is complete.
Q 12. A wire which is in N-S direction is dropped freely.
Will any potential difference be induced across its
ends ?
Ans. No, because neither horizontal component H nor
vertical component V of earth‘s magnetic field is
intercepted by the falling wire.
Q13 . A vertical metallic pole falls down through the
plane of magnetic meridian. Will any e.m.f. be induced
between its ends?
Ans. No, because the pole intercepts neither H nor V.
Q 14. A coil of metal wire is stationary in a non uniform
magnetic field. Is any e.m.f. induced in the
coil ?
Ans. No e.m.f. is induced as magnetic flux linked with the
stationary coil is not changing.
Q 15. Name the S.I. units of magnetic flux and magnetic
induction.
Ans. Weber. Tesla.
Q16. Name various methods of producing induced
e.m.f.
Ans. (i) By changing magnetic field B. (ii) By changing
area A in the magnetic field.
(iii) By changing relative orientation of the area w.r.t. the
magnetic field.
Q17. When is magnetic flux linked with a coil held in a
magnetic field zero ?
Ans. When plane of coil is along the field.
Q18. The induced e.m.f. is sometimes called back e.m.f.
Why ?
Ans. This is because induced e.m.f. opposes the current due
to the actual source of e.m.f.
Q19. Why are oscillations of a copper sheet in a
magnetic field highly damped ?
Ans. This is because of eddy currents developed in the
copper sheet.
Q20. What causes sparking in the switches when light
is put off?
Ans, Large induced e.m.f. at break causes the sparking.
Q21. What is the basic cause of induced e.m.f. ?
Ans. Change in magnetic flux linked with the circuit.
Q22. Does Lenz‟s law violate the principle of energy
conservation ?
Ans. No, Lenz‘s law does not violate this principle.
Q23. What is one henry ?
Ans. One henry is the self inductance of a coil in which a
current change at the rate of one ampere/sec, induces an
e.m.f. of one volt in the coil.
Q24. Are eddy currents useful or harmful ?
Ans. They are both, useful and harmful.
Q25. A closed loop of wire is being moved so that it
remains in a uniform magnetic field. What is the
current induced ?
Ans. Zero. This is because, when the loop remains in the
field, magnetic flux linked with the loop remains constant.
There is no change in magnetic flux.
Q26. What is EMI (electromagnetic induction)?
59
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Ans. EMI stands for electromagnetic induction. It is the
phenomenon of producing an e.m.f. in a circuit by changing
the magnetic flux linked with it.
Q27. Could a current be induced in a coil by rotating a
magnet inside the coil ? If so, how ?
Ans. Yes, by holding the magnet along the axis of the coil
and turning the magent about the diameter of the coil.
Q28. Can ever electric lines of force be closed curves ?
Ans. No, electric lines of force cannot be closed curves.
Q29. Is induced electric field conservative or non
conservative ?
Ans. It is non conservative.
Q30. Will an induced current be always produced
whenever there is a change in magnetic flux linked with
a coil ?
Ans. With change in magnetic flux, induced e.m.f. is a
must, but induced current will appear only when the circuit
is closed.
Q31. Can one have an inductance without a
resistance ? How about a resistance with an
inductance ?
Ans. No, as every material has some resistance. Yes, we
can coil a wire to have resistance with inductance.
Q32. A bar magnet M is dropped so that it falls
vertically through the coil C. The graph obtained for
voltage produced across the coil vs time
(i) Explain the shape of the graph
(ii) Why is the negative peak longer than the positive
peak ?
Ans. As the magnet M approaches coil C, magnetic flux
linked with the coil increases, emf. is induced in the coil,
which opposes the increase in flux. When magnet is inside
the coil, magnetic flux linked with the coil is constant.
Induced emf is zero. As the magnet falls below the coil,
mag. flux linked with the coil decreases, emf is induced in
the coil, which opposes the decrease in flux. As velocity of
magnet has increased, induced emf is more. Therefore,
negative peak is longer than the positive peak. When the
magnet has fallen through large distance, changing mag.
flux due to its movement vanishes. Induced emf reduces to
zero.
Q. 33. A cylindrical bar magnet is kept along the axis of
a circular coil and near it . Will there be any induced
emf at the terminals of the coil, when magnet is rotated
(a) about its own axis (b) about an axis perpendicular
to the length
of the magnet ?
Ans. (a) When the magnet is rotated about its own axis,
there is no change in magnetic flux linked with the coil.
Therefore, induced emf = 0.
(b) When the magnet is rotated about an axis perpendicular
to its length, orientation of the magnet wrt the coil
changes. Therefore, magnetic flux linked with the coil
changes . Hence an emf is induced in the coil.
Q 34. What is electromagnetic induction?
Sol.When the magnetic flux linking a conductor (or coil)
changes, an e.m.f. is induced in the conductor. If the
conductor (or coil) forms a complete loop or circuit, a
current will flow in it. This phenomenon is called
electromagnetic induction.
Q35. Define flux linkages.
Sol. The product of number of turns (N) of the coil and the
magnetic
flux (§) linking the coil is called flux linkages
i.e Flux linkages = N 
Q 36. What is Q-factor? Derive an expression of it.
Ans. Q-factor is that factor which measures the selectivity
or sharpness of a resonant
circuit.
Q
Voltage across L or C
Applied voltage
Q 37. Define wattless current.
Ans: Wattless current is that component of the circuit
current due to which the power consumed in the circuit is
zero.
5.7. ELECTRICAL MACHINES
(i)
Electric Generator:
(ii)
Transformer
1. Electric Generator:
An a.c. generator/dynamo is a machine which produces
alternating current
energy from mechanical energy.
Principle. An a.c. generator/dynamo is based on the
phenomenon of
electromagnetic induction i. e. whenever amount of
magnetic flux linked
with a coil changes, an e.m.f. is induced in the coil.
Construction. The essential parts of an a.c. dynamo are
shown in Fig.
60
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
1. Armature. ABCD is a rectangular armature coil. It
consists of a large
number of turns of insulated copper wire wound over a
laminated soft iron core. The coil can be rotated about the
central axis.
2. Field Magnets: N and S are the pole pieces of a strong
electromagnet
in which the armature coil is rotated. Axis of rotation is
perpendicular to
the magnetic field lines. The magnetic field is of the order
of 1 to 2 Tesla.
3. Slip Rings. S1 and S2 arc two hollowmetallic
rings, to which two ends of
armature coil are connected. These rings rotate with the
rotation of the coil.
Theory . As the armature coil is rotated in the magnetic
field, angle B
between the field and normal to the coil changes
continuously.
Therefore, magnetic flux linked with the coil changes. An
e.mf. is induced in the coil.
transformer is an electrical device used to increase or
decrease alternating voltage.
Types of Transformers
(i) Step-up transformer : The transformer which converts
low alternating voltage at higher current into a high
alternating voltage at lower current is called step-up
transformer. Inother words, a step up transformer gives
increased alternating voltage output.
(ii) Step-down transformer : The transformer which
converts high alternating voltage at current into a low
alternating voltage at higher current is called step-down
transformer. In other words, a step down transformer gives
decreased alternating voltage output.
Principle : A transformer is based on the principle of
mutual induction.
An e.m.f. is induced in a coil when a changing current
flows through its nearby coil.
d
dt
d
and Es   N s
dt
Es N s
Ns


, where
 k (transformation ratio)
Ep N p
Np
From Lenz ' s law, E p   N p
E.m.f. induced in the coil of generator:
 
Magnetic flux linked with coil ,   n ( B.A)
where n  number of turns of the coil .
Cases : (i) If k 1 then N s  N p , ( step  down transformer )
(ii) If k 1 then N s  N p , ( step  up transformer )
5.8. Types of energy losses in a transformer:
   nBA cos 
But   t
   nBA cos t
d
Since e  
( Lenz ' s law)
dt
d
 e   (nBA cos t )
dt
d
 e  nBA (cos t )
dt
d
 e  nBA ( sin t )
dt
 e  nBA sin t
 e  e0 sin t ,
where e0  nBA
Hence, induced e.m. f . e  e0 sin t
2. Transformer:
(i) Copper losses. Energy lost in windings of the
transformer is known as copper loss. Primary and
secondary coils of a transformer are generally made of
copper wires. These copper wires have resistance (R).
When current (I) flows through these wires, power loss
(I2R) takes place.
(ii). Flux losses. In actual transformer, the coupling
between primary and secondary coils is not perfect. It
means the magnetic flux linked with the primary coil is not
equal to the magnetic flux linked with the secondary coil.
(iii). Eddy Currents losses. When a changing magnetic flux
links with the iron core of the transformer, eddy currents
are set up. These eddy currents in the iron core produce
heat which leads to the wastage of energy. This energy loss
is reduced by using laminated iron cores.
Transformer is a device used to convert low alternating
voltage at higher current into high alternating voltage at
lower current and vice-versa. In other words, a
61
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Eddy currents are reduced in a laminated core because
their paths are broken as compared to solid core.
ADDITIONAL QUESTIONS
Q. Why is the core of a transformer laminated?
Ans:The core of transformer is laminated to reduce eddy
curents.
Q. Briefly explain the use of transformer in long
distance transmissions. Why is the long distance
transmission done at high voltage low current?
(Consider that 11000 watt of electric power is transmitted
first at 220 V and then at 22000 V.)
Ans. Loss of electric energy.
When transmission of
electric energy is done
at 220 V, a current of 50 A flows through the line wires[as
P = VI].
If R is resistance of line wires, energy equal to 502 R i.e.
2500 R joule will be dissipated per second as heat energy
[ as H=I2R for t=1 sec].
On the other hand, when transmission is done at 22000 V,
a current of 0.5A
only flows through the line wires.
In this case, the electric energy dissipated per second as
heat will be 0.52 R i.e. 0.25 R joule per second only.
Therefore, it may be concluded that if transmission is done
at high voltage, the transmission is much economical from
the point of view of the cost of line wires and poles.
Keeping in view the above two factors, transmission of
electric energy is done at high voltage and for this a step-up
transformer is used. But some part of the energy must be
lost as heat.
Q. A bulb connected in series with a solenoid is lit by
a.c. source. If a soft iron core is introduced in the
solenoid, will the bulb glow brighter ?
Ans. No, the bulb will glow dimmer. This is because on
introducing soft iron core in the solenoid, its inductance L
increases, the inductive reactance X L   L increases and
hence the current through the bulb decreases.
Q. How can you improve the quality factor of a series
resonance circuit ?
Ans. To improve the quality factor of series resonance
circuit, ohmic resistance of the circuit should be made as
small as possible.
Q. An electric lamp connected in series with a capacitor
and an a.c. source is glowing with certain brightness.
How does the brightness of the lamp change on
reducing the capacitance ?
Ans. Brightness of the lamp decreases. This is because on
reducing C, Xc increases, Z increases and I decreases.
Q.
Why cannot we use a.c, for electrolysis ?
.Ans. For electrolysis, we require fixed cathode and fixed
anode. In a.c. the direction of current is changing
periodically.
UNIT V (Electromagnetic Waves)
Therefore, we conclude that if transmission of electric
energy is done at high voltage the dissipation of_ energy is
much reduced.
KEY POINTS

2. Cost of transmission.
If transmission of electric
energy is done at 220 V, to transmit electric power of
11000 W, the current capacity of line wires has to be 50 A
and if transmission is done at 22000 V, the current capacity
of line wires has to be only 0.5 A. Therefore, if
transmission is done at low voltage, thick wires have to be
used. Due to the use of thick line wires, the cost of
transmission will increase. Further, stronger poles would be
needed in order to support thicker line wires. It will further
add to the cost of transmission. On the other hand, if the
transmission is done at high voltage low current the line
wires required are of low current capacity i.e. thin wires
and light poles may be used. Due to this, the cost, of
transmission will be low.

Electromagnetic waves: A transverse wave
consisting of time varying electric field and
magnetic field mutually perpendicular to each
other and perpendicular to the direction of
propagation.

In free space, electric field vector E and magnetic

field vector B are related by
C

E0 . where
B0
E0 and B0 are amplitudes of
electric field and magnetic field.
Expression for velocity of electromagnetic wave
is
1
C
0 E0
62
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
= 3108 ms .
where 0  4 107 TmA1
POWER QUESTIONS
0  8.854 1012 C 2 N 1m1






In a material medium c 
1

Displacement current (Id): Displacement
current is the current due to changing electric
field between the plates of a capacitor.
Mathematical expression of displacement current
is I  dE , where dE  changein flux
d
0
dt
dt
Need of displacement current: As suggested by
Maxwell, changing electric field intensity is the
cause of current through a capacitor. If this
concept is wrong, no current will flow through a
capacitor.
A charge oscillating harmonically is a source of
electromagnetic waves of same frequency. Eg.
An accelerated charged particle is a source of
electromagnetic waves.
Devices that produces electromagnetic waves are
LC oscillator, Antenna (which is equivalent to an
electric dipole).
To justify that electric dipole is source of
electromagnetic wave: An electric dipole consists
of positive and negative charges separated by a
distance of 2l. As the charges are oscillating, so
magnitude as well as direction of its dipole moment
changes with time. E.g. A transmission antenna.
5.9. Transverse nature of electromagnetic
wave:
As no ch arg e is enclosed by the parallelopipe
 
 E.dS  0
 
 
Therefore 
E.dS  
E.dS  0
ABCD
GEFG
( flux through the other faces have neglected )
 
 Ex S  Ex' S  0 [  E.dS   EdS cos 00 ]
As Ex' is antiparallel to Ex
 Ex S  E x' S  0
 Ex  Ex'
The above equation tells that electric field along x-axis
does not change ie. static.
so, Ex  E '  0 It means that the electric field is
x
Q380. Light can travel in vacuum whereas sound can
not do so. Why ?
Sol. Light being electromagnetic wave do not require any
material medium for its propagation. Hence light can travel
in vacuum. On the other hand sound is a mechanical wave.
It requires a material medium for its propagation. Hence
sound can not travel in vacuum.
Q381. State any four properties nf electromagnetic
waves.
Sol. The electromagnetic waves consist of wide range of
radiation, but all these radiations have the following
common properties.
1. They are transverse in nature.
2. They do not require any materia: medium for
propagation.
3. They travel with the same speed of 3 x 108 m/s in
vacuum.
4. They consist of mutually transverse varying electric and
magnetic fields.
Q. 382. Why does the electrical conductivity of earth‟s
atmosphere increase with altitude ?
Sol. We know, with an increase in altitude, the atmospheric
pressure decreases. The high energy particles (i.e. gammarays and cosmic rays) coming from outer space and
entering our earth‘s atmosphere cause ionisation of the atoms of the gases present there. The ionising power of
these radiation decreases rapidly as they approach to earth,
due to increase in no. of collisions with the gas atoms. It is
due to this reason that the elect rical conductivity of earth‘s
atmosphere increases with altitude.
Q383. charging current for a capacitor is 0 25A. What is
the displacement current across its plates?
Ans. 0.25 A, because magnitude of displacement current is
equal to that of conduction current.
Q. 384. Give difference between displacement current
and conduction current.
Ans. Conduction current is due to flow of electrons in the
circuit. It exists even if the flow of electrons is at uniform
rate.
Displacement current is due to time varying electric field. It
does not exist under steady condition.
Q.387. What oscillates in an electromagnetic wave ?
Ans. Electric and magnetic field vectors.
Q388. What features of electromagnetic waves led
Maxwell to conclude that light itself is electro- magnetic
wave ?
Ans. Transverse nature.
Q389. Arrange the following radiations in the
descending order of frequency: y-rays, infra-red rays,
microwaves, yellow light, ultraviolet rays.
Ans. gamma-rays, ultraviolet rays, yellow light, infrared
rays, microwaves.
Q390. Arrange the following radiations in the
descending order of the wavelengths : -rays, infrared
rays, red ligbt, yellow light, radio waves.
Ans. Radiowaves, infrared rays, red light, yellow light,rays.
perpendicular to the direction of the wave.
63
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
O391. From which layer of atmosphere Radio and
microwaves are reflected back ?
Ans. Ionosphere.
Q392. Which are the relevant waves in telecommunication?
Ans. Microwaves.
Q393. How are X-rays produced ?
Ans. When fast moving electrons are stopped suddenly on a
metal target of higher atomic number, then X-rays are
produced. The X-rays will also be produced when an
electron jumps from higher orbits to a vacancy on the inner
complete orbits in an atom of the element.
Q394. By which way the X-rays and  -rays of the
same energies can be distinguished ?
Ans. By the method of production.
Q395. Which has higher wavelength. rays, X-rays and
microwaves.
Ans. Microwaves.
Q396. Why are microwaves used in Radar ?
Ans. In a radar, a beam signal is needed in particular
direction which is possible if wavelength of signal wave is
very small. Since the wavelength of microwaves is a few
millimetre, hence they are used in radar.
Q297. Name the different layers of earth‟s atmosphere.
Ans. (i) Troposphere
(ii) Stratosphere (iii) Mesosphere
(iv) Iono-sphere.
Q298. What is the frequency range of the visible portion
of the electromagnetic spectrum ?
Ans. 4 x 1014 Hz ,to 8 x 1014 Hz.
Q 299. What are radiowaves ?
Ans. The electromagnetic waves of frequency ranging from
a few kilo hertz to about few hundred mega hertz
(i.e. wavelength 0. 3m and above) are called radiowaves.
Q300. Which part of the electromagnetic spectrum is
used in operating a RADAR ?
Ans. Microwaves.
Q301. Which of the following has shortest wavelength;
microwaves, ultraviolet rays and X-rays ?
Ans. X-rays
.Q302. Name the electromagnetic radiations used for
viewing the objects through haze and fog.
Ans. Infrared radiations.
Q303. Give the ratio of velocities of light rays of
wavelength 4000 A and 6000 A in vacuum.
Ans. One.
Q304. Why sky waves are not used in the transmission
of television signals ?
Ans. TV. signals are of high frequencies (100 MHz to 200
MHz). They can not be reflected to earth by ionosphere,
whereas sky waves are reflected from ionosphere. Hence
sky waves are not used for the transmission of TV signals..
Q305. What features of electromagnetic waves led
Maxwell to conclude that light itself is electromagnetic
wave ?
Ans. Light and electromagnetic waves are of transverse
nature and they travel with the same velocity in vacuum.
This led Maxwell to conclude that light itself is
electromagnetic wave.
Q306. Name the electromagnetic waves that have
frequencies greater than those of ultraviolet light but
less than those of gamma rays.
Ans. X-rays.
Q307 .The charging current for a capacitor is0.2A .
What is the displacement current ?
Ans. Displacement current = charging current = 0.2A.
Q 308 .What is the cause of conduction current ?
Ans. The cause of conduction current is the flow of
electrons in the conductor under the effect of potential
difference applied.
Q309. Which physical quantity, if any, has the same
value for waves belonging to the different parts of the
electromagnetic spectrum ?
Ans. The electromagnetic waves of different wavelengths
travel with the same speed (= 3 x 108 m/s)in vacuum.
Q310.What is the name given to that part of
electromagnetic spectrum which is used for taking
photographs of earth under foggy conditions from great
heights ?
Ans. Infrared rays.
Q311. Microwaves are used in Radar. Why ?
Ans. As microwaves are of smaller wavelengths, hence
they can be transmitted as a beam signal in a particular
direction much better than radiowaves because microwaves
do not bend around the corners of any obstacle coming in
their path.
Q 312 .State two applications of Infrared radiations.
Sol. Infrared radiations are used (i) to treat muscular strain
(ii) for taking photographs during the conditions of fog,
smoke etc.
Q 313. State two applications of Ultraviolet radiations.
Ans. Ultraviolet radiations are used (i) to preserve the food
stuff (ii) for sterilizing the surgical instruments.
Q 314. State two applications of X-rays.
Ans. X-rays are used (i) for the detection of fractures in the
bones of human body (ii) for the detection of explosives,
opium and gold in the body of the smugglers.
Q315. Show that electromagnetic wave is produced
when charge is accelerated.
Ans. When charge is accelerated, both the electric and
magnetic fields produced will change with space and time.
These fields are perpendicular to each other. Due to which
electromagnetic wave is produced. This means that an
accelerated charge emits electromagnetic wave.
Q 316. For long distance Radio broadcast, we use short
wave band only. Why ?
Ans: Because ionosphere reflects the waves lying in
frequency range of short wave band.
Q317.Explain the „Green House Effect of earth‟s
atmosphere.
Ans. Our earth‘s atmosphere-is transparent to the visible
radiations coming from sun, stars etc. but reflects back the
infrared radiations and hence it does not allow the infrared
radiations to pass. The energy from the sun, heats the earth
which in turn starts emitting radiations. Since the earth gets
heated to much lower temperature than trie temperature of
sun, the radiations emitted by earth are mostly in the
infrared region, according to Planck‘s law. These radiations
emitted by earth are reflected back by earth‘s atmosphere.
Due to which the earth‘s surface remains warm at night.
This phenomenon is called Green House effect.
Q 318 Explain the terms, (i) ground wave and (ii) sky
wave.
64
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Ans. (i) The amplitude modulated radio waves which are
travelling directly following the surface of the earth are
known as ground waves or surface waves. These waves can
have frequency upto 1500 k Hz (or wave length more than
200 m). The ground waves can bend round the corners of
the objects on earth and hence their intensity falls with
distance.
(a) The amplitude modulated radio waves of frequency
more than 1500 k Hz (or wavelength below 200 m) which
are received after being reflected from the ionosphere are
called as sky waves.
Q319. What is ground wave ? Why short wave
communication over long distance is not possible via
ground waves ?
Ans. The amplitude modulated radiowaves having
frequency upto 1500 k Hz (or wavelength more than 200
m) which are travelling directly following the surface of
earth are known as ground waves. The short wave
communication over long distance is only possible via sky
waves. It is not possible via ground waves because the
ground waves can bend round the corners of the objects on
earth and hence, their intensity falls with distance.
Moreover the ground wave transmission becomes weaker
as frequency increases, due to conduction losses.
Q320. Is it necessary to use satellite for long distance
T.V transmission ? Give reason.
Ans. The T.V. transmission involves the television signal
waves having the frequency range 80 MHz to 200 MHz.
These waves neither follow the curvature of earth nor they
get reflected by ionosphere. Therefore their communication
via sky wave is not possible. The reception of television
signals is possible either (i) by using communication
geostationary satellite which reflects the television signals
back to earth or (ii) by using tall receiver antenna which
may directly intercept the signals.
Q321. Explain that microwaves are better carriers of
signals than radio waves.
Ans. Microwaves are the electromagnetic waves of
wavelength of the order of a few millimeters, which is less
than those of TV. signals. On account of smaller
wavelength, the microwaves can be transmitted as beam
signals in a particular direction, much better than
radiowaves because microwaves do not spread or bend
around the corners of any obstacle coming in their way.
Therefore microwaves are better carriers of signals than
radio waves.
tall receiver antenna which may directly intercept the
signals.
Q322.Why sky wave propagation of electromagnetic
waves cannot be used for T.V. transmission ? Suggest
two methods by which range of T.V. transmission can
be increased.
Ans. The sky wave propagation deals with amplitude
modulated microwaves which can be reflected by
ionosphere of earth‘s atmosphere and the frequency of
these radio waves are more than 1500 k Hz (or wavelength
below 200 m). The T.V. transmission deals with frequency
range 80 MHz to 200 MHz. Their communication via sky
wave is not possible, since they are not reflected back to
earth by ionosphere.
The range of TV. transmission can be increased
(i) by using communication geostationary satellite which
reflect the television signals back to earth or (ii) by using
Q. 328. Give one use of each of the followings
(i) Infrared rays (ii) Gamma rays (iii) microwaves (iv)
Ultraviolet rays.
Ans. (i) Infrared rays are used in physical therapy i.e. to
treat muscular strain.
(ii) Gamma rays are used in the treatment of cancer and
tumors.
(iii) Microwaves are used in Radar system for air craft
navigation.
(iv) Ultraviolet rays are used to destroy the bacteria and the
sterilizing the surgical instruments.
Q323. What does an electromagnetic wave consist of?
On what factors does its velocity in vacuum depend?
Ans. Electromagnetic waves are those waves in which there
are sinu soidal variation of electric and magnetic field
vectors, at right angles to each other as well as at right
angles to the direction of its propagation. These two field
vectors vibrate in the same phase and with the same
frequency.
Q324. How would the following be affected in the
absence of atmosphere around the earth ? (a) Surface
temperature of earth (b) Range of radio waves
transmission. Give brief reason for your answer in each.
Ans. (a) Temperature of earth would be lowered because
the green house effect of the atmosphere would be absent,
(b) Range of radio waves transmission will decrease
because there will be no reflecting layer for radiowaves as
are there in ionosphere.
Q325. “Greater is the height of TV transmitting
antenna greater is its range”. Prove.
Ans:
Range of TV cov ering d  2 Rh
i.e., d  h
Q. 326. Which of the following, if any, can act as a
source of electromagnetic waves ? (i) A charge moving
with a constant velocity (ii) A charge movng in a
circular orbit (iii) A charge at rest. Give reason.
Ans. A charge moving in a circular orbit can produce
electromagnetic waves
because circular motion is an
accelerated motion and accelerated charge produces
e.m.waves.
Q. 327. Identify the part of electromagnetic spectrum,
to which waves of frequency (i) 1020 Hz (ii) 109 Hz
belong.
Ans. (i) 1020 Hz. corresponds to y- rays and (ii) 109 Hz.
corresponds to microwaves.
Q329. When can a charge act as a source of
electromagnetic waves ? How are the directions, of the
65
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
electric and
magnetic field vectors, in an
electromagnetic wave, related to each other and to the
direction of propagation the wave ?
Ans. Only an accelerated charge can act as a source of
electromagnetic waves. The directions of electric field
vector and magnetic field vector are perpendicular to each
other as well as perpendicular to the direction of
propagation uf the electromagnetic wave.
360
 odd , then number of images is given

by 360

(viii)
If an object approaches or moves away from
a plane mirror with a velocity of ‗v‘ then
the image will approach or move away with
a speed of ‗2v‘
Spherical Mirror
Spherical mirror is a part of a reflecting spherical surface.
Types of spherical mirror:
(i) Concave Mirror: It is a part of a hollow sphere having
outer part i.e. bulging surface) silvered and the inner part
(i.e. depressed surface) as a reflecting surface.
Uses: Used by dentists, eye specialists to focus light on
teeth, or eyes, in the ear, driver‘s mirror.
(ii) Convex Mirror: It is a part of a hollow sphere having
inner part (i.e. depressed surface) silvered and outer part
(i.e. bulging surface) as a reflecting surface.
Uses: Used as reflectors in cinema projectors, as objective
in reflecting type telescopes, as reflectors in solar cookers.
EXAM QUESTIONS
1.
2.
3.
UNIT VI
Name the type of mirror which has focal
length equal to infinity.
Ans. Plane mirror.
What is radius of curvature and focal length of
a plane mirror?
Ans. R=infinity and f=infinity.
Using Cartesian sign conventions, derive the
mirror formula for a convex mirror.
Ans.
OPTICS
KEY POINTS
Nature of the image formed by a plane mirror
(i) Image is at the same size as that of the object.
(ii) Image distance=Object distance
(iii)
Image formed is erect.
(iv)
Image is laterally inverted (left and right
alternation. It means that the left part of the
object will appear as the right part and viceversa)
(v)
If the height of object is h, to see the full
length image, length of mirror must be h/2
(vi)
For rotation of θ by the mirror, the reflected
ray turns through 2θ
(vii)
If θ is the angle between two inclined plane
mirror, then number of images formed is
360
is an even integer. If
n
 1 if

66
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Here  ' s ABC and A ' B ' C are similar

A ' B ' CA '


           (i )
AB
CA
Similary  ' s ABP and A ' B ' P are similar

A ' B ' PA '

           (ii )
AB
PA
Comparing equations (i ) and (ii ), we get

CA ' PA '
PC  PA ' PA '



  (iii )
CA
PA
PC  PA
PA
U sin g sign conventions we get ,

PC  R, PA '  v; PA  u
EXAM QUESTIONS
Now equation (iii ) becomes,
Rv
v

R  u u
  uR  uv  vR  uv
1.
What is meant by refraction? What is its
cause?
Ans: The phenomenon of bending of light from
its path when it travels from one medium to
another medium is called refraction of light.
The cause of refraction is that speed of light is
different in different media.
 vR  uR  2uv
Dividing on both sides by uvR we get ,
1 1 2
 
v u R
But
R 2f
1 1 1

 
f
v u
The minimum distance between the real object
and real image formed by a lens is 4f, where f is
the focal length of the lens.
The focal length of a convex lens immersed in a
liquid (whose refractive index is greater than the
refractive index of the glass) becomes negative.
That is, convex lens behaves as a diverging lens
(or concave lens) is such a liquid.
The power of a lens immersed in water decreases
and becomes 1/4 times its original power.
Q2. Define (i) Refractive index and (ii) Snell‟s law
Ans: (i) Refractive index n  sin i or n  c
sin r
Q4. Explain why is a convex mirror used as driver‟s
mirror?
Ans: The image produced by a convex mirror is erect and
diminished. As the size of the image is small, the field of
view is wide and the driver can see a wider area.
Q5. Can a plane mirror form an inverted image?
Ans: If the object is placed vertically above a mirror (the
mirrior is kept horizontally on ground ), the image formed
will be inverted.
f=
c

KEY POINTS









Intensity of image  Aperture 2
The brightness of image decreases when a lens is
half painted however the size of the image
remains unaffected.
If a lens is cut into two equal halves horizontally,
intensity becomes 1/2th.
If the lens is cut into two equal halves vertically,
intensity and aperture remains the same.
Refractive index of a medium is always positive.
During refraction of light, wavelength and
velocity of light changes but the frequency
remains unchanged.
When object is in denser medium, then the
apparent depth is "less than actual depth, if
observed from the rarer medium.
When object is in rarer medium, then apparent
depth is greater than the actual depth, if observed
from the denser medium.
Full size image of an object is seen even if half of
the portion of the lens is blocked or painted
black. However, intensity or brightness of the
image decreases.
v
(ii) According to Snell‘s law sin i  n1
sin r n2
Q3. Light of wavelength in air enters a medium of
refractive index 1.4. What will be its frequency in the
medium?
Ans:
0
3 108 

1A=1010 m 
5000 1010 

 6 1014 Hz

Q4.Can absolute refractive index of any medium be less
than one?
Ans.
No
c
Absolute refractive index n=
v
Where c  speed of light and v  speed of light in the medium
Since v  c, n  1
Hence n cannot be less than one
Q5. What is critical angle?
Ans: The angle of incidence at which angle of refraction
becomes 900 is known as the critical angle.
Q6. What is optical fibre? What is its main use? What is
its principle?
Ans: Optical fibre is thin and long strands of glass coated
with a transparent medium of lower refractive
index.Optical fibre is mainly used for transmitting light
from one place to another without any loss in the intensity
of light.
67
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
An optical fibre works on the principle of total internal
reflection.
Q7. Is light signal transmitted through optical fibre
100% efficient?
Ans: Practically not. the intensity of light gradually
decreases when the light signal is transmitted through long
distances as the glass consists of impurities present in the
core and the cladding.
Q8. What is the relation between the refractive index
and
the
critical
angle?
Ans: The refrative index and the critical angle are related as
n
1
sin C
5.10. Conditions for Total Internal Reflection
(i) The light should travel from denser to rarer medium.
(n) The angle of incidence must be greater than the critical
angle for the given pair of media.
Q17. Show that 1n  1
2
sin c
Ans.
sin i
2
Since, n1 
angle of refraction =900 )
 2 n1 
sin c
 2 n1  sin c
sin 900

Q9. What is the power of a lens? What is one Dioptre?
Ans.

 1
1
1 
 ( n  1) 


f
 R1 R2 
 1
1 
 P  ( n  1) 


 R1 R2 
sin r
If , i  c; r  900 (i,e angle of incidence = critical angle,
1
1
1
 sin c  1n2 
n2
sin c
Q18. Derive lens maker‟s formula.Ans. It is called lens
maker‘s formula because of its application while
manufacturing lenses. Assumptions: (i) The lens is thin
and have small aperture.(ii) Object is a point source lying o
the principle axis.
Q10. An object is held at the principal focus of a
concave lens of focal length F. Where is the image
formed ?
Ans.
Between
focus
and
optical
centre.
Q11. A diverging lens of focal length '/ is cut into two
identical parts each forming a plano-concave lens. What
is the focal length of each part ?
Ans. 2f
Q12. A glass lens of refractive index 1.5 placed in a
liquid. What must be refractive index of the liquid in
order to make the lens disappear ?
Ans. 15 i.e. same as that of glass.
Q 13. A converging lens of refractive index 1.5 kept in a
liquid medium having same refractive index. What
would be the focal length of the lens in this medium ?
Ans. Will disappear.
Q. 14. How does the power of a concave lens vary, if the
incident red light is replaced by violet light ?
Ans. Power increases because focal length is minimum for
violet light.
Q. 15. How does refraction of light affect the length of
the day ?Q. 16. What is total internal reflection ? Under
what conditions does it take place ?
Or
What do you mean by total internal reflection of light ?
What are essential conditions for it?
Ans. The phenomenon of reflection when a ray of light
travelling from a denser to rarer medium is sent back to the
same denser medium provided it strikes the interface of the
denser and the rarer media at an angle greater than the
critical angle is called total internal reflection.
Let n1  refractive index of air (rarer )
n2  refractive index of material of the lens (denser )
.
Step1: Re fraction through XPY
1 ,

n1 n2
n  n1

 2
u
v1
R1
       (1)
Step 2 : Re fraction through XP2Y ,
( hereI1 is the object and I is the image)

n2
n
n  n2
 1  1
v1
v
R2
       (2)
Adding equations (1) and (2) we get ,
n1 n2  n2  n1
n  n1 n1  n2


 2


u
v1  v1  v
R1
R2
 1
n1 n2 n2
n1
1 




  n2  n1  


u
v1
v1
v
 R1 R2 
 1
n
n
1 
 1  1   n2  n1  


v
u
 R1 R2 

 1
 1
1
1 
 n1 
    n2  n1  


u
v
 R1 R2 
1
1  n2  n1   1
1 

 



v
u
n1
 R1 R2 
 n2
 1
1
1   1
1
1

 1 


 
 
f
v
u
 n1
  R1 R2   f
 1
1
1 
  n  1 


f
 R1 R2 
19. If the wavelength of incident light on convex lens is
increased, how will the focal length change?
68
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Ans.
Here,
1
1
1
1
 (n  1) ( 
) or
n
f
R1 R2
f
n
and
1
2
f  2

Hence, the focal length increases with the increase in
wavelengthof light.
20. A thin converging lens has focal length f, when
illuminated by violet light. State with reason, how the
focal length of the lens will change,if violet light is
replaced by red light.
Ans.
1  r1  i  1  i  r1
 2  r2  e   2  e  r2
In  PBC ,  is the exterior angle
   1   2 (exterior angle =
sum of opposite interior angles)
 (i  r1 )  (e  r2 )
  (i  e)  (r1  r2 )      (1)
Ans.
For a lens f 
Since n 
red  violet
1
2
1
n
.
 f  2
 A  O  1800 (opposite angles of quadrilateral ABOC)
 f red  f violet .
 r1   r2  O  1800 (sum of three angles = 1800 )
5.11.PRISM

A  O   r1   r2  O

A   r1   r2
 A  r1  r2        (2)
1.
 From equation (1),   i  e  A
Plot graph to show the variation of the angle
of deviation as a function of angle of incidence
for light rays passing through the prism.
Ans
 A  i  e
sin(
5.
Prove that
n 
A m
2
A
sin( )
2
)
Ans.
Here,   A  i  e.
At min imum deviation position,
i  e, r1  r2  r ( say ),    m (angle of min imum deviation)
 m  A  i  i
2.
.
Write down the relation between the refractive
index of the material of the prism, angle of
prism and angle of minimum deviation.
A  m
sin(
)
Ans.
n
3.
4.
2
A
sin( )
2
What do you mean by dispersion of light?
Ans. The phenomenon of splitting of white light
(i.e., polychromatic light) into its constituent
colours is called dispersion of light.
Show that A    i  e
 2i  A   m
A  m
.
2
Also A  r1  r2
i 
 A  r  r[ r1  r2  r ]
 2r  A
r 
A
.
2
 From Snell ' s law, n 
A  m
)
2 .
A
sin( )
2
sin(
Ans.
Q. Explain the cause of dispersion using
Cauchy‟s formula.
69
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)

Ans.
From Cauchy ' s formula n  A 
B
2
 ........

Where A, B are cons tan ts.
1
or n  2 .

Since red  violet , nred  nviolet

Also for a thin prism,   (n  1) A.
 red  violet
or  violet  red . [as   n]
6.
How would the sky look in the absence of
atmosphere?
Ans:
No atmosphere means no scattering of
light. So the sky would be perfectly black in
colour.You could have observed twinkling of
stars during day time as the sky would be
perfectly dark.
7. What is scattering of light? State Rayleigh‟s
law of scattering.
Sol. It is the phenomenon in which light incident on very
small molecules of air is radiated in all directions.
Ans. According to Rayleigh‘s law of scattering, the
intensity of the light scattered is inversely proportional to
the fourth power of the wavelength. ie.
I
1
4
Explain the Sun appears reddish at Sunset or
Sun-rise
Ans. At sunset or sunrise, the sun and its surrounding
appears less scattering of blue colour red because of the
scattering of light.
The light from the sun at sunset or sunrise travels a longer
distance through the earth‘s atmosphere to reach our eyes.
This light is deprived of blue colour due to large scattering
of blue light by the particles present in the atmosphere and
rich in red colour. So the sun appears reddish at sunset or
sun-rise because the red colour is least scattered.
9. Explain the blue Colour of Sky.
Ans. Blue colour of sky is due to scattering of light. The
wavelength of blue colour is much smaller than that of red
1
I 4
colour. From Rayleigh‘s principle, , ie.
5.12.Sustained Interference
To have a sustained interference, the following conditions
must be fulfilled-:
1. Two sources must be coherent sources of light.
2.
Two sources of light should emit light waves
continuously.
3. Two coherent sources of light should be close to each
other.
4. The amplitude of the two waves emitted by the two
coherent sources of light should be equal to get complete
contrast between dark and bright bands.
5. The two coherent sources of light should be very
narrow.
6. The distance of the screen from the coherent sources of
light should not be small.
7. The light should be monochromatic.
EXAM QUESTIONS
8.

intensity of scattering of blue colour is much more than that
of red colour. Due to this reason, blue colour becomes the
major colour and the sky appears to be blue in colour.
HUYGENS PRINCIPLE AND INTERFERENCE
KEY POINTS


Law of conservation of energy is not violated in
the interference of light. The light energy absent
at the dark fringes appear at the bright fringes.
Two waves are said to be in the same phase if
the phase angle between them is zero.
The interference pattern disappears if one of the
slits is closed as interference requires two
coherent sources of light.
If the monochromatic source is replaced by
white light
(i)
the fringes are coloured (ii) the
central bright fringe is white
A monochromatic source is a light source with a
single wavelength. Eg. red.
1.
Define wavefront of light.
Ans. Wavefront: A line or a plane on which all the particles
equidistant from the source vibrating in the same medium
are located.
Or
Locus of all the particle of the medium vibrating
in the same medium is called wavefront.
2.
What is the shape of a wavefront emitted by a
light source in the form of narrow slit?
Ans. Cylindrical wavefront.
3.
Write the different types of wavefronts and their
sources.
Types of wavefront:
(i) Spherical wavefront ( produce be a point source)
(ii) Cylindrical wavefront (produce by a line source)
(iii) Plane wavefront (produce by a point source
when the wavefront is considered at a large
distance from the point source).
Q. State Huygen‟s principle of secondary
wavelets.
Ans. (i) Every point on a given wavefront acts as a source
of secondary wavelets that move in the forward direction at
the same speed at which the wave moves.
(ii) The new position of the wavefront at any instant is a
line drawn tangent to the edges of the wavelets at that
instant.
70
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
(iii) The rays are perpendicular to the wavefront.
Limitation:The obliquity factor 1  cos   gives
the intensity of the secondary wavelets. Along
backward direction,   1800 . So, (1-1)=0 as
sin 1800=-1. It means that thre is no flow of
energy when light waves travel in forward
direction. However it is not correct.
4.
What is the phase difference between two
points on a wavefront?
Ans. Zero.
5.
What is meant by interference of light? Write
the uses and applications of interference.
Ans. Interference pattern is said to be sustained
if the positions of constructive interference (i.e.
bright fringes) and destructive interference (i.e.
dark fringes) remain fixed on the screen.
Uses.
1. Phenomenon of interference is used for the
precise determination of the wave length of light.
2. It is used to determine refractive index or thickness of
transparent sheets.
3. It is used to test the flatness of plane surfaces.
4. It is used to design optical filter which allows a narrow
band of wave length to pass thr
5. It is used in holography to produce 3-D images.
Formation of fine colour layers on oil film, colouring
effect produced by soap bubbles in sunlight, colouring
effects produced by birds and peacock etc are due to
interference of light.
Q. What are coherent sources of light? What
about incoherent sources?
Ans. Two sources of light are said to be coherent if they
emit waves of same frequency (or wavelength) and are
either in phase or have a constant initial phase difference.
Ans. The sources having different frequencies or the
sources having same frequency but no stable phase
difference are known as incoherent sources.
9. How is the intensity and amplitude of a wave related?
Ans: Intensity  (amplitude) 2
10. In Young's double slit experiment, the distance between the
slits is halved, what change in the fringe width will take
place?
D
Ans:
Here,  =
d
D
d
'
. Since d '  .
d'
2
 D 2 D
 ' 

d
d
2
 2 .
5.13.Interference Fringes
Dark and Bright fringes in the interference pattern are
called interference fringes.
Fringe width(  ): The distance between any two
consecutive dark or bright fringes is called fringe width.
To find the expression for fringe width:
Assume that the light waves are emitted by coherent sources S1 and S2 .
From right angled  S2 PB,
2
6.
State the path difference between two waves
for (1) constructive interference
(2)destructive interference.
Ans.
(1) The path difference = n
(2) The path difference = (2n  1)

 d
S2 P 2  S2 B 2  PB 2  D 2   x            (1)
2

and from triangle S1PA,
2
 d
S1 P 2  S1 A2  PA2  D 2   x            (2)
 2
As path S2 P is greater than path S1P,
path difference x  S 2 P  S1P
2
71
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Subtracting equation (2) from (1), we obtain
i.e, x  (2n  1)

d 
d 



S 2 P 2  S1 P 2  D 2   x     D 2   x   
2 
2






d d2  2
d d2 
2
2
2
 D  x  2x 
 D  x  2x 

2
4 
2
4 
2
 D2  x2  2 x
2
d d2
d d2

 D2  x2  2 x 
2
4
2
4
 xd  xd
 2 xd
Since S 2 P 2  S1 P 2   S 2 P  S1 P
 S P
2
 S1 P

 a  b   a  b  a  b  
2
2
  S 2 P  S1 P
  S2 P
 S P  S P   2 xd
2 xd
S P 
S
P
S P 

2
1
xd  (2n  1)

D
2
 (2n  1) D
x 
2d
If n  1, x1 
D
, (1st dark fringe)
2d
3 D
If n  2, x2 
, (2nd dark fringe)
2d
2 D  D  D
 Fringe width (  )  x2  x1 


.
d
d
d
( In terms of bright fringes )
3 D  D 2 D  D



2d
2d
2d
d
( In terms of dark fringes )
or ,   x2  x1 
1
Since x  S 2 P  S1 P

EXAM QUESTIONS
2 xd
x 
S
P
 2  S1P

1.
As the po int P is very close to central bright fringe O,
On what factors does the fringe width depend
upon?
S1 P  S 2 P
Ans: Fringe width  
Also, S1 and S 2 are very close to each other ,
 S 2 P  S1 P  D  D  2 D
S P
2
2 xd
 S1 P

 x 
ie. path difference x 
2 xd xd

2D
D
Bright fringes will be formed if the path difference equals
an integral multiple of wavelength.
ie. x  n
; where n  0,1, 2,3.....................
xd
so, n 
D
n D
x
d
If n  0, x0  0 (central bright fringe)
If n  1, x1 
slit and screen.
(3) inversely proportional to the width of slit.
xd
D
Bright Fringes (constructive interference)
D
(1st bright )
d
2 D
If n  2, x2 
(2nd bright )
d
......................................................
Dark Fringes: For obtaining Dark Fringes, path difference
must be equal to odd multiple of half wavelength.
D
d
The fringe width is (1) directly proportional to
the wavelength of monochromatic light.
(2) directly proportional to the distance detween
D  S1 P  S 2 P
 x 
2

1
2

2.
What is the change in fringe width if the whole
apparatus of Young‟s Double Experiment is
completely immersed in a liquid?
Ans.
Here  
D
(in air )
d
If wavelength   wavelength in air , wavelength in liquid
will be given by   

where n  refractive index of
n
medium of liquid .
Since n1 (refractive index for any medium is always
greater than 1)
  
Since there is reduction in  , fringe width  decreases
as   
Obtain the expression for fringe width when
the whole apparatus of Young‟s experiment is
completely immersed in a liquid.
Ans.
Here  
D
d
(before immersion)
 D
 
( after immersion)
d
No change in D and d (except  )
     wavelength in the medium;



n  n  refractive index of the medium 
D

  

nd
n
Also   
72
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
D
.
d
(1) When screen is moved closer, D decreases and hence 
decreases.
(2) when separation between the slits increases, d increases and
hence  decreases.
Since,  
KEY POINTS



If two slits are illuminated with two different
basic colours say red and blue, they will no
longer act as coherent sources.There will be no
interference pattern on the screen.
The central bright in interference pattern may be
bright if the abstacle and its image on reflection
act as coherent sources.
If the distance between slit and screen D is made
very large the fringe width will be too large and
the whole screen will be occupied by either bright
or dark fringe i,e interference will not be
observed.
NEW QUESTIONS FOR HSE - 2011
Q1. Define wavefront of light.
Q2. What is the shape of wavefront emitted by a (1)
narrow slit? (2) a point source?
Ans: (1) Cylindrical (2) Spherical.
Q3 State Huygen‟s principle of secondary wavelets.
Q4. What is the phase difference between two points on
a wavefront?
Ans: Zero.
Q5. What are coherent sources of light?
Q6. State the conditions that must be satisfied for two
light sources
to be
coherent.
Ans: (1) They must emit light waves continuously of same
wavelengths.
(1) The phase difference between these waves must be
zero.
(2) Explain why two identical but independent
sources of light cannot be coherent sources.
(3) Ans: Even though the sources are identical, the two
sources cannot produce waves in the same
phase.Hence, the two independent sources cannot
act as coherent sources.
Q. How does the fringe width of interference change if the
whole apparatus of Young's experiment is immersed in a
liquid of refractive index 1.3?
Ans: If  = fringe width in air.
 '  fringe width in the liquid.
then  ' 

n


1.33
.
Q12.Can two coherent sources of light be obtained from
one non-coherent source of light? Explain.
Ans: It is possible, In Young‘s experiment, two coherent
sources are obtained from the double slit when light from a
source fall on it.
Q13.What is the effect on the interference pattern in
Young‟s double slit experiment when (1) screen is move
closer to the slits? (2) separation between the slits is
increased?
Ans:
Q. If the coherent sources are far apart, will you detect
interference?
Ans: Here,  =
D
.
d
If the coherent sources are far apart, d increases and
corresponding value of  decreases to large extent.Since
fringe width becomes very small interference pattern will
not be observed.
Q. In Young‟s double slit experiment, three colours
green, yellow and red are successively used. For which
colour the fringe width will be maximum?
D

i, e   . Since  for red colour is max imum, so
d
fringe width corresponding to red colour will be max imum
Q17.State the reason why two independent sources of
light cannot be considered as coherent.
Ans: Because they cannot maintain same phase relationship
throughout the travel of their light energies.
Q18.What are the conditions for the sustained
interference?
Ans: (i) The sources of light must be coherent. (ii) The
coherent sources must emit light waves continuously. (iii)
The amplitudes of waves from the coherent sources must
be the same to have good contrast between dark and bright
fringes.
Q19.Using Huygens‟ principle, draw a diagram to show
propagation of a wavefront originating from a
monochromatic point source.
Ans. See Figure (you look by yourself)
Q20.Is it possible to obtain two coherent sources of light
from one non-coherent source?
Ans: As seen in Young‘s double slit experiment, two
coherent sources of light can be obtained from a noncoherent source.
Q21.In Young‟s double slit experiment one slit is
covered so that no light enters. How will the intensity of
light in the region of central maximum vary? Explain.
Ans:
The intensity of light varies directly to the square of amplitude.
i.e, I  (Amplitude) 2
It means that the amplitude is the result of the two sources.If
one of the slit is closed,amplitude in the region of central maximum
is halved.
Hence the intensity becomes one-fourth.
73
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Q22.Give two examples illustrating the phenomenon of
interference of light.
Ans: (1) Interference pattern produced by Young‘s double
slit experiment.
(2) Colouring effects produced by the thin oil film or
the water surface by soap bubble.
Q23. In Young‟s double slit experiment, if the
monochromatic source is replaced by white light, how is
the interference observed?
Ans. The central bright fringes for all colours are at the
same position. Therefore, the central bright fringe is white.
Q24. Is the law of conservation of energy hold true in
interference of light?
Ans. Yes, the energy destroyed at the dark fringes are
obtained at the bright fringes.


KEY
Two independent sources (non-coherent) cannot
produce interference since the light waves
produced by the source cannot have same phase.
If the coherent sources are widely apart, the
fringe width will be infinitely small. It means that
the dark and bright fringes will be too close to
each other and hence interference pattern might
not be observed.
5.14.DIFFRACTION AND
POLARISATION
The phenomenon of bending of light around the corners of
an obstacle or an aperture into the region of geometrical
shadow of obstacle is called diffraction of light.
Pronounce diffraction
Diffraction os said to be more pronounced when size of the
obstacle is of the order of the wavelength.
Fresnel Diffraction. In this type of diffraction, the source
or the screen or both are at finite distances from the
obstacle or the aperture causing the diffraction . The
phenomenon of diffraction obtained on the screen is known
as Fresnel diffraction pattern.
Fraunhoffer diffraction. In the Fraunhoffer diffraction,
the source and the screen are at infinite distance from the
obstacle or the aperture causing the diffraction.
BN  path difference
tan  
x
D
 (1)
BN
 BN  d sin 
d
For min ima, path difference  n
and sin  
 n  d sin 
for 1st min ima, n  1
  d sin   s in  

 (2)
d
Since  is very small , sin   tan 
From equations (1) and (2)
x 
D
   D  xd  x 
D d
d
Since width of central max imum  x  x  2 x
2
D
d
EASY SCORE
Width of central maximum increases if
(i) The distance between screen and slit is separated.
(ii) Width of slit os decreased.
(iii) Wavelength of light increases.
Unpolarised Light:
Light consisting of electric field vectors vibrating in every
plane perpendicular to the direction of propagation is called
unpolarised light.
Polarisation of light:
The phenomenon of restricting light vector to vibrate in a
plane perpendicular to the direction of propagation is called
polarisation of light.
Plane of vibration: The plane containing the
crystallographic axis in which vibrations occur.
Plane of polarisation: The plane perpendicular to the
plane of polarisation.
Polaroid: A device used to produce polarisation.
Analyser: A polaroid used to detect or examine the
polarised light is called analyser.
Width of central maximum
The distance between the first two minima on the screen is
known as width of central maximum.
5.15.To show the transverse nature of light:
When two polaroids are parallel to each other, then the
light transmitted by the first polaroid is also transmitted by
the second polaroid as shown in Figure A.
74
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
When the two polaroids are crossed i.e. they aremutually
perpendicular to each other, then the light transmitted by
the first polaroid is completely blocked by the second
polaroid as shown in Figure B.
When the axis of the crystal A is parallel to the direction of
the vibrations of the plane polarised light , this plane
polarised light is transmitted as such by the crystal A.
When the axis of the crystal A is perpendicular to the
direction of vibrations of the plane polarised light , then the
vibrations of the plane polarised light is completely
blocked. Thus, there is no light transmitted by the crystal A
and hence the intensity of light transmitted is zero. The
crystal A identifies the polarisation of light and hence it is
known as analyser.
This experiment further shows that the light is transverse in
nature.
EXAMINATION QUESTIONS 2011
Q1.Which phenomena establish the wave nature of light
?
Ans. . These are interference, diffraction and polarisation of
light.
Q2. In Young‟s double slit experiment, if S1 and S2 are
illuminated by two bulbs of same power, what will be
observed on the screen ?
Ans. No interference pattern will be observed on the
screen. This is because waves reaching at any point on the
screen do not have a constant phase difference, as phase
difference from two incoherent sources changes randomly.
Therefore, maxima and minima would also change their
positions randomly and in quick succession. This will result
in general illumination on the screen.
Q3. Two slits in Young‟s double slit experiment are
illuminated by two different sodium lamps emitting
light of same wavelength. Do you observe any
interference pattern on the screen ?
Ans. No, interference pattern is not obtained. This is
because phase difference between the light waves emitted
from two lamps will change continuously.
Q4. Why is interference pattern not detected, when the
two coherent sources are far
apart ?
Ans. Because fringe width   1
d
therefore, when d is so large, the width may reduce beyond
the visible region. Hence the pattern will not be seen.
Q5. No interference pattern is detected when two
coherent sources are very close to each other.
Why?
Ans. When d is negligibly small, fringe width which is
proportional to 1/d may become too large. Even a single
fringe may occupy the screen. Hence the pattern cannot be
detected.
Q6. Out of speed, frequency and wavelength, name the
parameter (s) which remain the same on refraction.
Ans. Only frequency remains the same on refraction.
Q7 . Why cannot we obtain interference using two
independent sources of light ?
Ans. This is because two independent sources of light
cannot be coherent, as their relative phases are changing
randomly.
Q8. If a wave undergoes refraction, what happens to its
phase ?
Ans. No phase change occurs during refraction.
Q9. In Young‟s double slit experiment, the intensity of
central maximum is I. What will be the intensity at
the same place if one slit is closed ?
Ans. When one slit is closed, amplitude becomes I/2 and
hence Intensity becomes l/4th, and there is no interference.
Q10. What will be the effect on the fringes, if Young‟s
double slit experiment set up is immersed in water ?
Ans. The fringes become narrower.
Q11. Does interference of light give information about
longitudinal/transverse nature of light waves ?
Ans. No, interference is only a wave phenomenon.
Q12. Two waves of amplitudes 3 mm and 2 mm reach a
point in the same phase. What is the resultant
amplitude ?
Ans. Resultant amplitude 3+2 = 5 mm.
Q13.Can two independent sources of light be coherent ?
Ans. No.
Q14. What happens to interference pattern when one of
the slits in Young‟s double slit experiment is closed?
Ans. The interference pattern disappears.
Q15. Widths of two slits in Young‟s experiment are in
the ratio 4 : 1. What is the ratio of the amplitudes of
light waves from them ?
Ans.
W1
I
a2
4
 1  2 
W2
I2
b
1

a
2

 a : b  2 :1
b
1
Q16. What are coherent sources of light ?
Ans. The sources of ligftf which emit light waves of same
wavelength, same frequency and in same phase or having a
constant phase difference are called coherent sources.
Q17. Oil floating on water looks coloured due to
interference of light. What should be the approximate
thickness of the film for such effects to be visible ?
Ans. For interference effect to be visible, thickness of oil
film must be of the order of wavelength of visible light,
which varies from 4000 A to 8000 A.
75
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)

EASY SCORE


Colour on thin oil film appears to be dark if the
thickness of the film is large.
Our eye cannot detect polarised and unpolarised
light.
5.16. Uses of polaroids and plane polarised light
(i) Polaroids are used in sun glasses to avoid glare produced
by shiny surfaces.
(ii) Polaroids are used trains and aeroplanes as window
glasses.
(iii) Polaroids are used to record and reproduce three
dimensional moving pictures.
(iv) Polaroids are used in microscopes to see very minute
particles which are not able to see due to glare.
(v)
The principle of polarisation is used in
Liquid Crystal Display (L.C.Ds) in
calculators, TVs, computer monitors etc.
5.17.Brewster‟s law
Red light of wavelength 6500 A from a distant source falls on a slit
0.50 mm wide.What is the distance between the two dark bands on
each side of the central bright band of diffraction pattern observed
on a screen placed 1.8 from the slit?
Ans: Distance between the two dark bands on each side of
the central maxima of
diffraction pattern = width of the central maxima.
 x
2 D
.
d

Here,   6500 A  6.5 107 m
D  1.8m
d  0.5mm  0.5 103 m
x 
2  6.5 107 1.8
 4.8 103 m
0.5 103
KEY POINTS



If the lenses are absent in Fraunhoffer diffraction
( like in fresnel diffraction), the central maximum
may be dark.
Electromagnetic waves such as X-ray,
Radiowave and sound wave etc. also show
diffraction pattern.
Crossed polaroids refers the combination of two
polaroids kept such that their axes are
perpendicular to each other.
EXAMINATION QUESTIONS -2011
Q1. Complete polarisation is produced at a glancing
angle of 370 . Calculate the refractive index.
Ans. From Brewster‘s law,
glancing angle  angle of incidence  900
 370  i  900  i  900  370  530
 n  tan  p  tan 530  1.32
Brewster‘s law states that the refractive index of a
refracting medium is equal to the angle of polarisation.
Q2. What do you meant by diffraction of light?
Ans. Bending of light around the corner of obstacle is
called diffraction of light.
Proof:
Q3. What is the order of the size of obstacle to observe
diffraction?
Ans. To produce diffraction, size of obstacle must be of the
order of the wavelength.
n  tan  p
BOY  COY  900
 (1)
r  BOY  900
 (2)
r   COY  900
 (3)
from equations (1) and (2),
BOY  COY  r  BOY
 r  COY
Q4. Why are short waves used in long distance
broadcast?
Ans. Short waves are diffracted less and hence can be
transmitted as a beam.
from equation (3) r   r  900
 r   900  r
sin i
Since n 
. For complete polarisation
sin r 
i   p  angle of polarisation.
n
sin  p
sin(900  r )

sin  p
cos r

sin  p
 i   p  r; laws of relection 
cos  p 
 tan  p
 n  tan  p ,
Q5.Write the expression for the width of central
maximum of diffraction pattern produced by a single
slit.
Ans.
Q6. Differentiate between interference and diffraction.
Ans. (i) Interference arises due to the superposition of
waves from two coherent sources. However, diffraction
pattern is produced as a result of waves coming from
different parts of the same wavefront.
which proves Brewster ' s law.
76
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
(ii) In interference pattern, all the bright fringes are of the
same intensity. However, in diffraction pattern, all bright
fringes are not of the same intensity.
(iii) In interference pattern, the width of the fringes may or
may not be the same. On the other hand, diffraction fringes
are not of the same width.
(iv) In interference pattern, the points of minimum
intensity are perfecdy dark. However, in diffraction pattern,
the intensity at minima is never zero.
Q7. Define plane of vibration.
Ans: Refer yourself.
Q8. Define plane of vibration.
Q9.
“Unpolarised light is symmetrical about the
direction of propagation of light”. Comment.
will you show experimentally the light waves are
transverse in nature ?
What are polaroids ? Write their four uses.
(a) Two near by narrow slits are illuminated by a
single monochromatic source. Name the
patterns obtained on the screen, (b) One of the
slits is now covered. What is the name of the
pattern now obtained on the
(b) screen ?
Ans.(a) Interference pattern (b) Diffraction pattern.
Q10. What is polarisation of light ? What is its most
important significance ?
Q11. What is plane polarised light ? Name any two
methods to produce plane polarised light.
Ans. Plane polarised light can be produced by reflection
and scattering of light.
first four parts of the slit send wavelets in opposite phase
and so on.
Q16. What two main changes in diffraction r pattern of
a single slit will you observe when the monochromatic
source of light is replaced by a source of white light ?
Ans. When the monochromatic source of light is replaced
by the source of white light, the following changes in
diffraction pattern of a single slit occur :
(i) The diffraction bands are coloured, although the central
maxima or band is white. (ii) Since band width is directly
proportional to wave length, so the width of red band is
wider than the width of blue band.
Q17. If we look at the sun through a piece of fine
muslin cloth, what do we observe and why ?
Ans. Muslin cloth is made up of fine threads and the space
between these threads behaves as fine slit. When the sun
light (having light of all colours) falls on these slits, the
diffraction of light takes place. As a result of this, coloured
streaks are is observed.
Q17.What is the cause of luminous border surrounding
the profile of a mountain just before the sun rises or
sets ?
Ans. This is due to diffraction of light.
Q18. Light waves can be polarised but the sound waves
cannot be polarised. Explain why ?
Ans. We know that only transverse waves can be polarised,
whereas longitudinal waves cannot be polarised. Since light
waves are transverse in nature and the sound waves are
longitudinal in nature, therefore, only light waves can be
polarised.
Q12. State and prove Brewster‟s law.
Q19.
Which of the phenomena established the
transverse nature of light ?
Ans. The phenomenon of the polarisation of light
established the transverse nature of light.
Q13. What is meant by plane polarized light ? What
type of waves show the property of polarisation ?
Describe a method for producing a beam of plane
polarised light.
Q20. Can our eyes distinguish between polarised and
unpolarised light ?
Ans. No, our eyes cannot distinguish between polarised and
unpolarised light.
Ans. Light in which the vibration of electric field vector is
restricted in a plane is known as plane polarized light.
Q21 . What happens to the intensity of light when it is
polarised ?
Ans. When the light is polarised, the vibration of electric
vector are restricted only in one plane. It means the
vibrations of electric vector in other planes are cut off.
Hence intensity of the polarised light is less than, that of the
unpolarised light.
Q14.Why does the intensity of secondary maxi-mum
becomes less than as compared to the central
maximum?
Ans. Intensity of central maximum of diffraction pattern is
due to the wavelets of all parts of the slit exposed to the
light. However, intensity of first secondary maximum is
due to the wavelets from only one-third part of the slit as
the first two parts of the slit send wavelets in opposite
phase. The intensity of second secondary maximum is due
to the wavelets from only one-fifth part of the slit as the
5.18. HUMAN EYE
Important Terms:
(1) Accomodation of eye: Formation of sharp image at the
retina due to fine adjustment by cilliary muscles.
(2) Near point: The nearest point from the eye at which the
eye can see a sharp image. It is 25cm for a normal eye.
77
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
(3) Far point: It is the farthest point from an eye at which
an object can be placed so that a sharp image is formed. For
a normal eye it is infinity.
(4) Power of accomodation: It is the maximum variation
in the power of the eye lens.
Correction: To correct short-sighted vision, a diverging
lens (concave lens) of
suitable focal length is placed in front of the eye. The rays
of light from distant
object are diverged by the concave lens so that final image
is formed at the retina.
(ii) Farsightedness (or Hyper-metropia): A person who
can see distant objects clearly but cannot focus on near
objects is farsighted. Whereas the normal eye has a near
point of about 25 cm,
a farsighted person may have anear point several metres
from the eye.
This defect may occur if the diameter of person‘s eyeball is
smaller than the usual or if the lens of the eye is unable to
curve when ciliary muscles contract. In such a case, for an
object placed at the normal near point (i.e., 25 cm from
eye), the image of the object is formed behind the retina as
shown in Fig.
5.19.Eye Defect
The inability of an eye to see near and far distances clearly
is known as defect of vision.
Types of defects:
(i) Short sightedness (or Myopia): A person who can see
the near objects clearly but cannot focus on distant objects
is short sighted. The far point of a short-sighted person may
be only a few metres rather than at infinity. This defect
occurs if a person‘s eyeball is larger than the usual
diameter. In such a case, the imageof a distant object is
formed in front of the retina as shown in Fig.
It is because the eye‘s lens remains too converging,
forming the image of the object in front of the retina.
Correction: It can be corrected by using a convex lens of
correct focal length.
(iii) Presbyopia i.e. loss of power of accommodation.
In this defect both far off and nearer objects are not clearly
seen. This defect is
generally corrected by using bi-focal lenses.
(iv) Astigmatism. In this defect, focal length of eye lens in
two orthogonal directions vary which makes it difficult to
clearly see the object in two said directions simultaneously.
This delect due to improper spherical eye lens is corrected
by using cylindrical lens in specific direction.
An astigmatic eye cannot focus on both horizontal and
vertical lines simultaneously.
POWER QUESTIONS
1. If the far point of a myopic eye is 150cm, calculate the
power and focal length of the concave lense to be used
for this correction.
78
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Ans.
Here, u   (ve as the dis tan ce is measured from the eye)
d  150cm
1
1
1
1
From the formula,   , we get

or f  150cm.
f
d
f
150
100
100
Power of lense is P  

 0.67 D
f (in cm)
150
Prove that for a normal eye, the power of accomodation
is about 4 dioptre(D).
The dis tan ce of the retina from the eye lens is about
2.5cm.Therefore, image dis tan ce is v  2.5cm
Object dis tan ce is u  25cm(ve as measured from the eye against
the rays )
1
1 1
 
f
u v
1
1
1
1 11





25 2.5 2.5 25 25
25
 f  cm.
11
100
11
Power of eye lens P 
 100   40 D
f (cm)
25
And the far po int of normal eye is . So u   and v  2.5cm

U sin g
1
1 1 10
  
f
u v 25
25
 f  cm
10
25
 40 D
10
Maximum var iation in power of eye lens  44 D  40 D  4 D
Hence power of eye  lens Pfar  100 
5.20. COMPOUND MICROSCOPE
Principle: When an object is placed in front of a convex
lens of small focal length at a distance between F and 2F,
the real, inverted and magnified image is formed on the
other side of this lens. If this image lies within the focal
length of another convex lens E of large aperture then the
image acts as an object for this lens. The final image
produced by this lens is virtual, inverted and highly
magnified.
Magnifying Power :
The magnifying power of a compound
microscope is defined as the ratio of the angle
subtended by the final image at the eye to the
angle subtended by the object at the eye when
both are at a distance of least distance of distinct
vision.
5.21. MAGNIFYING POWER (OR ANGULAR
MAGNIFICATION)
The magnifying power of a compound microscope is
defined as the ratio of the angle subtended m the final
image at the eye to the angle subtended by the object at the
eye when both are at a distance of least distance of distinct
vision.
Here α = angle subtended by the object at the eye when
placed at the least distance of distinct vision. and
β = angle subtended by the final image at the eye when
placed at distance of distinct vision.

M .P 

tan 
Since  and  are very small M .P 
tan 
AB
AP
and tan  
C2 A
C2 A
AB
tan  C2 A AB
 M .P 


AP
tan 
AP
C2 A
tan  
but AP  AB
AB
 M .P 
. Now multiplying on both sides by AB 
AB
AB AB AB AB
M .P 



AB AB AB AB
image height image dis tan ce
Since magnification (m) 

object height object dis tan ce
AB
AB
 me 
and for objective mo 
AB
AB
 M .P  me  mo
v
AB vo
Now mo 
  o  from sign convention 
AB uo uo
AB ve
me 
 (both are  ve and they cancel )
AB ue
79
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Since ve  D  me 
D
ue
The magnifying power is given by M .P 
Applying lens formula we have,
sin ce the angles are small , M .P 
1 1 1
 
f e ve ue

1
1
1


Now, multiplying on both sides by D
f e  D ue

D D
 1
ue f e
 M .P 
and C2 A  ue ( from sign convention)
vo  D 
  1
uo  f e

Special case: If the object is very close to the
focus of the object lens,
the image AB is formed very close to the eye lens.
Now vo  L
 M .P  
AB
AB
and  
C2 A
C1 A
AB
tan  C2 A C1 A
 M .P 


. S in ce C1 A  f o
tan  AB C2 A
C1 A
Then, tan  
1
1
1


 ve   D and ue  ue 
f e  D ue
D
D
 1 
fe
ue
tan 
tan 


 fo
ue
The value of M .P depends on the adjustment of the telescope
(i ) distinct vision and (ii ) normal vision
1 1 1
(i ) distinct vision :
 
But from sign convention
f e ve ue
u  ue and ve   D
L
D
1  
f0 
fe 

POWER QUESTION
Q. Among the lenses of power 3D, 7D, 10D and 15D,
which lenses will be choosen for constructing a
compound microscope?
Ans.
1
Since magnifying power M .P  Pe Po 
 M .P 
fe fo
For high magnifying power, the power of lenses must be
large so lens of 10D and 15D must be selected. Since object
lens has small focal length, its power must be higher than
the eye lens. Hence, lens of 15D must be used as object
lens and lens of 10D must be used as eye lens.
5.22. TELESCOPE
Magnifying
power
(Angular
magnification)
of
Astronomical telescope.
Magnifying power of an astronomical telescope is defined
as the ratio of the angle subtended the final image at the eye
to the angle subtended by the object at the eye.
1
1
1
1 1


 
f e  D ue ue D
f 
1
1 1 1
   1  e 
ue f e D f e  D 
f 
f 
 M .P  o 1  e  . In this case, dis tan ce between the
fe  D 

lenses  length of tube
 L  C1 Fo  C2 Fo  f o  ue
EASY SCORE
If the image is to seen brightly, the whole light must enter
the eye. It means that the aperture of eyepiece must be
small.
The aperture of the objective must be large to increase
magnifying power by gathering more light.
ADDITIONAL SCORE ON TELESCOPE
A good telescope should have high magnifying power, high
resolving power and large light gathering power.
(a) Magnifying power (M.P.) :
The magnifying power of a telescope for normal vision is
given by,
M .P 
 fo
fe
. M.P. will be large if the focal length
of an eyepiece is very small while the focal length of an
objective lens is very large.
(b) Resolving power is given by R . P

D
1.22 
D is the diameter of the objective lens and is the
wavelength of light used.
The telescope should have the objective lens of large
diameter.
The resolving power is better if object is illuminated by a
light of small wavelength.
(c) Light gathering power. The light gathering power (or
brightness) of a telescope is
 r
2

D2
4
where D is the diameter of the objective.
80
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
A telescope will have large light gathering power if the
diameter of the objective lens is large. Hence a bright
image will be formed by the telescope.
Limitations
The refracting type telescope (using lenses) suffers from
spherical and chromatic aberrations.
Due to these aberrations, the final image of the object is
coloured and blurred. Hence the minute details of the object
cannot be analysed properly.
The objective lenses of very large aperture are very
difficult to manufacture.
Advantages and Disadvantages of Reflecting Type
Telescope
Advantages
(i) Free from chromatic and spherical aberrations. Hence
sharp image of the object is formed.
(ii) Mirror weighs much less than a lens of similar quality.
(iv) Due to good reflection the image formed by these
telescope is quite bright.
(iv) The paraboloidal mirrors of large aperture can be easily
manufactured.
Disadvantages
(i) Need frequent adjustments and hence inconvenient to
use.
(ii) They cannot be used for general purposes.
6.0. Reflecting Type Telescope
A reflecting type telescope uses a paraboloidal mirror of
large aperture which is free from abberations.
Types:
(i) Cassegrain Type Telescope. It consists of a concave
mirror O of large aperture with a circular hole in its centre,
A small convex mirror is placed in front of the objective O
of the telescope. The final image is observed through an
eye piece placed in front of the hole of the objective.
(ii) Newtonian Telescope. It consists of a concave mirror
of large aperture known as objective. This mirror is fitted at
one end of the metallic tube whose other end is open. A
plane mirror M is placed at an angle of 45° with the axis of
the tube. The final image is seen through the eyepiece fitted
at one side of the tube as shown in Figure.
The magnifying power of a reflecting telescope is given by
M .P 
fo
fe
Difference between Refracting type telescope and
Reflecting type telescope
Refracting Type Telescope
1. The objective is achromatic converging lens.
2. It suffers from chromatic and spherical aberrations and
hence image formed is coloured and blurred.
3. The light gathering power is small, so a faint image of
the distant star is observed.
4. It is used for general purpose and is handy.
Reflecting Type Telescope
1. The objective is paraboloidal mirror.
2. It is free from chromatic and spherical aberrations,
so image is sharp and bright.
3.
The light gathering power is large, so a bright image
of the distant star is observed.
4. It is used in astronomy and is not handy.
6.1. Resolving Power of Optical Instruments
Important terms:
Meaning of un-resolved and resolvedIf two close objects cannot be distinguished by the optical
instrument, they are said to be unresolved whereas they are
said to be resolved if the images are well distinguished.
Limit of resolution:
The minimum distance of separation between two points so
that they are seen as separated or just resolved by the
optical instrument is known as limit of resolution.
81
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Resolving Power of optical instrument:
It can be defined as the ability of an optical instrument to
form distinctly separate images of the two very vlose
objects.
It is also defined as the reciprocal of limit of resolution.
Resolving Power of human eye R.P  D
; where
1.22
D=diameter of pupil of eye.
Resolving power of astronomical telescope R.P  D ;
1.22
where D= diameter of the object lens.
Larger is the diameter of the object lens, more will be the
resolving power.
Resolving
power
of
compound
microscope
2n sin  ;
R.P 
1.22
where = numerical aperture and =wavelength of light
used.
EASY SCORE
The resolving power of compound microscope can be
increased by increasing the refractive index n of the
medium. In oil-immersion microscopes, it is achieved by
filling the medium between object and objective with a
transparent oil of higher refractive index.
Ans. The telescope forms the image of the distant object
near the eye. So the angle subtended by the image to the
eye is large, hence magnifying power of telescope
increases. As a result of this, magnified image of the distant
object will be seen through the telescope.
Q4. Reflecting type telescope is preferred in astronomy.
Explain why ?
Ans Astronomy is the study of heavenly bodies which are
very far away from us. To analyse these objects, we require
an optical instrument which may produce a bright and
sharp image of these objects. Since reflecting type
telescope is free from chromatic and spherical aberrations,
so a sharp image of distant heavenly object like star can be
formed by this telescope, [moreover, the aperture of the
objective of this telescope is very large and it collects large
amount of light and hence a bright image is formed by this
telescope.
Q5. To increase the resolving power of a refracting type
telescope, objective lens should be of very large
aperture. Then why can‟t be make the aperture as large
as possible ?
Ans. The large aperture will increase the resolving power
of the telescope but it also increases the spherical
aberration which mars the quality of the image.
Q6. The resolving power of a microscope increases
when red light illuminating the object is replaced by the
blue light. Explain why ?
Ans. R.P. of microscope ;
R. P 
1

when wavelength length decreases, the resolving power
increases. (wavelength of blue light is less than that of red
light).
EXAMINATION QUESTIONS-2011
Q1. Very distant stars not visible to the eye are visible
through telescope. Explain, why ?
Ans. Due to small aperture of eye it collects less amount of
light not sufficient to excite the retina whereas telescope
with large object lens collect more light which is sufficient
to excite the retina.
Q2. The magnifying powers of two telescopes are same
but the apertures of their objectives are different. What
will be the difference in the final images formed by
them ?
Ans. The brightness of the images will be different as it
depends upon the diameter of the aperture. The final image
formed by the telescope having an objective of large
aperture will be more bright. Moreover, the resolving
power R.P 
D of that telescope will be more than the
1.22
telescope having an objective of smaller aperture.
Q3.
Distant object which appear quite small with
naked eye, appear larger through the telescope.
Explain why ?
Q7. „Telescope resolves where as microscope
magnifies‟. Explain.
Ans. Magnification compares the size of the image with
that of the object as done by microscope whereas
resolution deals with the fact whether images of two nearby
objects are distinguished or not as done by telescopes.
Q8. What is presbiopia? How is it corrected?
Q9. Define limit of resolution of a telescope.
Q10. Define
microscope.
resolving
power
of
a
compound
Q11. How is the resolving power of a telescope change
by increasing or decreasing the aperture of the
objective?
Ans: . Resolving power
R.P 
D
1.22 
increases if the
diameter increases and vice-versa.
Q12. Using a ray diagram, show the image formation in
a compound microscope. What is the nature of the
image formed ?
82
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Q13. The objective and eyepieces used in a
compound microscope are not single lenses but are the
combinations of more than one lenses placed in contact
with each other. Explain why ?
Ans. To minimize abberations.
Q14. Draw a labelled ray diagram showing the
formation of image in a compound microscope. Also
define the magnifying power of a compound
Q15. What is the magnifying power of a telescope whose
objective and eye-piece have focal lengths 180 cm and 3
cm respectively ?
Ans. M .P  f o =180/3=60.
fe
Q16. Define the terms : magnifying power and „revolving power for a telescope .
Q17. What is the effect of increasing the diameter of the
objective of a telescope on its (i) Magnifying power and
(ii) resolving power ?
Ans. (i) M.P. remains unchanged (ii) R.P. increases with
increase in D.
Q18. Draw a labelled ray diagram to show the
formation of the image of a distant object through a
reflecting telescope ?
Unit VII: Dual Nature of Matter
and Radiation
Photoelectric emission: Metals consists of free valence
electrons which are responsible for electrical conduction.
These free electrons cannot leave the surface of metals due
to surface barrier. The electrons can leave the surface only
when external energy is supplied to the electrons. This
process of emission of electrons from the metal surface is
known as photoelectric emission.
Work function (Wo) : The minimum energy required
by the free electron to leave the metal surface without
moving in space is known as work function.


Work function is measured in electron volt (eV)
1eV=1.6×10-19C×1V=1.6×10-19J
Photoelectric effect: The process of emission of
photoelectrons from the metal surface when light
of suitable frequency greater than threshold
frequency falls on the surface is known as
photoelectric effect.
Q19. Draw a labelled ray diagram to show the
formation of image in a refracting type astronomical
telescope. Why should the diameter of the objective of a
telescope be large ?
Q20. Draw the course of rays in an astronomical
telescope, when the Final image is formed at infinity.
Also define the magnifying power of the astronomical
telescope in this position.
Q21. Write two advantages of the reflecting type
telescope over a refracting type telescope. On what
factors does its resolving power depend ?
Ans. R.P. depends upon (i) the aperture of the concave
mirror and (ii) the wavelength of light used.
Q22. How is the resolving power of a telescope change
by increasing or decreasing the aperture of the
objective ?
Raman effect: Raman scattering or the Raman effect is
the inelastic scattering of a photon. It was discovered by Sir
Chandrasekhara Venkata Raman and Kariamanickam Srinivasa
Krishnan in liquids.
When light is scattered from an atom or molecule, most photons
are elastically scattered (Rayleigh scattering), such that the
scattered photons have the same energy (frequency) and
wavelength as the incident photons. However, a small fraction of
the scattered light (approximately 1 in 10 million photons) is
scattered by an excitation, with the scattered photons having a
frequency different from, and usually lower than, the frequency
of the incident photons.
6.2. Hertz and Lenard experiment:
In Hertz‘s experiment, electromagnetic waves were
produced with the help of induction coil(IC).
Electromagnetic waves were produced across spheres S 1
and S2. Hertz observed that sparks across S1' S2 ' of detector
jumped more readily when detector was exposed to ultra
violet light from an arc lamp. This observation led him
conclude that light favoured the emission of some electrons
from the spheres S1 and S2. Phillip Lenard observed that
when UV light fell on a metallic electrode (cathode) ,
electric current appeared in the circuit and the current
developed was called photoelectric current. When the UV
light was not exposed to the cathode photoelectrons were
not emitted.
83
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
(ii)
The number of photoelectrons emitted per
second by a substance is directly
proportional to the intensity of incident light
provided incident light has a frequency
greater than threshold frequency.
The maximum KE of photoelectrons emitted
ins directly proportional to the frequency of
the incident light provided the intensity of
incident light is greater than the threshold
frequency.
The process of photoelectric emission is
instantaneous. (It means that photoelectrons
are emitted as soon as light is incident on the
surface without any delay)
(iii)
(iv)
Threshold frequency ( o):The minimum frequency of
incident light required to emit electrons from a metal
surface is known as threshold frequency.
Stopping potential (Vo): It is the
minimum negative
potential applied to anode plate for which photoelectric
current becomes zero is known as stopping potential.

Photoelectric effect cannot be explained on the
basis of wave theory of light.
6.4. Einstein‟s Photoelectric equation:
According to Einstein, the energy supplied by the photon is
used by the electron in two ways: (i) as work function to
overcome the surface barrier (ii) as KE of emitted
photoelectrons.
Saturation current: When all the photoelectrons emitted
by cathode reach the anode plate, the photoelectric current
becomes maximum and is known as saturation current.





Maximum KE of photoelectrons
2
of
photoelectron is directly proportional to the
stopping potential.
The intensity of incident radiation does not affect
the stopping potential.
For f>fo (frequency of incident light>threshold
frequency) , the stopping potential (Vo) is directly
proportional to the frequency of incident light.
Value of saturation current does not depend upon
the frequency of the incident radiation.
Threshold frequency will be more for materials
having higher work function.
6.3. Laws of photoelectric effect:
(i)
When light of suitable frequency greater
than threshold frequency is incident on a
metal surface photoelectrons are emitted.
Energy from photon=Work function (W0)+KE
Since E  h and KE 
1
mv 2 max
2
1
mv 2 max    (i )
2
When    0 ; no photoelectrons are emitted i.e. KE  0
h  W0 
 h 0  W0  0 which gives W0  h 0
Thus from eq.(i ) h  h 0 

1
mv 2 max
2
1
mv 2 max  h  h 0  h   0 
2
Which is Einstein‟s photoelectric equation.
Application of Einstein‟s photoelectric equation: (To
verify the laws of photoelectric equation using Einstein‘s
photoelectric equation)
84
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
FromEinstein ' s equation,
1
mv 2 max  h   0 
2
1
mv 2 max   ve(i.e. photoelectrons are emitted )
2
1
(ii ) If    0 ;
mv 2 max  ve(i.e. photoelectrons are not emitted )
2
(i ) If    0 ;
De-Broglie equation
Energy of photon is E  h
1
1
Since  
 c
; where c  speed of light in vacuum.

 E  hc

1

Since E  mc 2
giving hc
Hence, photoelectrons are not emitted when frequency of
incident light is less than the threshold frequency. On the
other hand, photoelectrons are emitted when the frequency
of incident light is greater than the threshold frequency.
Photon: Photon is a packet of energy. It is also defined as
quantum of energy ejected at the speed of light. Energy of
photon is given by E  h ; where h=Planck‟s
constant(=6.63×10-34Js) and  =frequency of radiation.
Properties of photon: (i) Photon travel with the speed of
light (ii) Photons are electrically neutral. It means that
photons are not deflected by electric and magnetic fields.
(iii) Photons travel with the speed of light. (iv) Wavelength
of photon changes in different media and hence velocity of
phton is different in different media.
6.5. To estimate the momentum of photon:
Energy of photon is E  h
1
1
Since      c ; where c  speed of light in vacuum.


1
 E  hc

1
h
Since E  mc 2 giving hc  mc 2   mc


h
h
 

mc p
where p  mc(momentum of photon)
De-Broglie waves: When a body moves with a velocity, it
radiates waves known as de-Broglie waves. The
wavelength of this wave is known as de-Broglie
wavelength.
As suggested by Louis de-Broglie, a French physicist, all
the particles such as electrons, protons, neutrons etc. have
dual nature. It means that all material particle can behave
both as wave as well as particle.
1

 mc 2 
h

 mc
h
h

mc
p
where p  mc (momentum of photon)
 
Conclusions: (i) de-Broglie wavelength is inversely
proportional to the velocity of the particle and to the mass
of particle ( 

h
mc
) (ii) If the particle is at rest (c=0), the
de-Broglie wavelength is infinite and hence these waves
cannot be observed. (iii) For a particle other than photon,
the velocity will be v and hence



 
h
mv
de-Broglie wavelength has no relation with the
charge on the moving body.
de-Broglie waves are not electromagnetic in
nature as they are not produced by charged
particles.
de-Broglie waves are also not mechanical waves
as they can travel through free space. These
waves are probability waves.
6.6. Expression for de-Broglie wavelength of an electron
moving through a potential difference: When theelectron
moves through the potential difference it gains a K.E.
1
K .E gained by electron  mv 2 and energy possessed by electron
2
when accelerated by potential difference V is E  eV
1
Comparing , mv 2  eV ; next step is to multiplied by m on both sides
2
1 2 2
 m v  meV  m2v 2  2meV
2
mv  2meV
h
h 

 mv  2meV 
mv
2meV 
Substituting h  6.63 1034 Js; m  9.11031 Kg (mass of electron)
As de  Broglie had suggested ,  
e  1.6 1019 C
 
6.63 1034
31
19
2  9.110 1.6 10 V

12.27 1010
m
V
85
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
6.7. DAVISSON AND GERMER EXPERIMENT

The main objective of the experiment was to study wave
nature of electrons. In the experiment, a beam of slow
electrons from an electron gun is made to fall on a nickel
crystal. The intensity of scattering was measured by a
detector fitted on a graduated scale.
It was observed that when a beam of electrons accelerated
through a potential difference of 54 volts was made to
incident on the nickel crystal, the intensity of scattering was
found to be maximum at scattering angle of 500.
6.8. Experimental verification of Davisson and
Germer experiment



Photoelectrons are the electrons emitted by a
metallic surface when light of suitable frequency
falls on the surface.
1eV is the K.E gained by an electron when
moving through a potential difference of 1 volt.
1eV=1.6×10-19J
The phenomenon which illustrates the particle
nature of light is the photoelectric effect.
Numerical problem
Q1. A photon has energy of 5.2×10-20J. Calculate its
frequency.
Ans:
As we know, E  h
Here E  5.2  1020 J ; h  6.63  1034 Js
Then,  
E
5.2  1020

 0.78  1014 Hz.
h
6.63  1034
Q2. Photoelectric work function of a metal is 1eV. If
0
light of wavelength 3000 A falls on it, calculate the
velocity of the ejected electron.
Here,       1800
Ans:
 2  1800  
Work function , W0  1eV  1.6 1019 J
scattering angle   500 ; giving 2  1800  500
  65
0
0
  3000 A  3000 1010 m  3 10 7 m
From Bragg ' s equation for max ima,
2d sin   n
As E  h 
hc 
1
1

  ;  c where c  cons tan t 
 



 E  h 
hc
0
as n  1, d  0.91 A ( for nickel )
0
 2  0.91 sin 65   ; giving   1.82  0.9063 A
0


6.63 1034  3 108
 6.62 1019 J
3 107
0
 1.65 A (exp erimental )
From de  Broglie hypothesis,  

12.27 0
A
V
0
12.27 0
A  1.67 A (theoritical )
54
Since the two results are in close agreement with each
other, it confirms the de-Broglie hypothesis of wave
nature
of
moving
particles.
Conceptual points

One photon is capable of ejecting one
electron
 De-Broglie waves are associated with
moving charged or uncharged bodies Visible
light can cause photoelectric emission on
alkali metals.
 . Ultraviolet light can cause photoelectric
effect on metals.
1
mv 2  h  W0  Einstein ' s equation 
2
 mv 2  2  h  W0   2  6.62  1019  1.6  1019 
Since KE 
 v2 
v
13.24 1019  3.2 1019
m
10.4 1019
 1106 ms 1
9.11031
Q3. Calculate the de-Broglie wavelength of an electron
beam accelerated through a potential difference of 4V.
Ans: Here, V  4volt.
 
0
12.27 0 12.27 0
A
A  6.135 A
2
4
Examination Questions
86
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Q4. The wavelength  of a photon and the de-Broglie
wavelength of an electron have the same value. Show
2 mc times the K.E of
h
that energy of the photon is
But  
c

h
 1 
2m
c

For a photon, E  h 
 (i )

h
h
; Squaring gives 12 
2mc
2mc
 2mc  2
Hence,   
 1
 h 
electron, where m, c and h have their usual meanings.
hc

 mv 
1
1 m2v 2
mv 2 

2
2 m
2m
h
Since  
( de  Broglie wavelength )
mv
2
K .E 
2
h
 
h
h2
1

mv 
so, K .E     2 

2m

2m
Dividing (i ) by (2),
 (2)
hc
E
hc 2m 2
2m c
 2 



h
1
K .E

h2
h

 2 2m
 2m c 
 E  K .E 

 h 
Which proves the required condition.
Q5. An  particle, a proton and electron having same
K.E have wave nature also. Compare their de-Broglie
wavelength.
Since  
h

2meV
Q8. A photon and electron have got same de-Broglie
wavelength. Which has greater total energy?
1
m
Hence m  m p  me giving    p  c
Ans:
i.e. wavelength of alpha particle is the least and that of
electron is maximum.
Q6. An electromagnetic wave of wavelength  is
incident on a photosensitive surface of negligible work
function. If the photoelectrons emitted from this surface
de-Broglie
1 ,
wavelength
prove
that
 2 mc  2
 

 h  1
As
Since  
E  W0  K .E ( Einstein ' s equation)
 K .E 
h
W0
But W0  1( neglected )
h
2m  KE 
; squaring gives  2 
h2
( KE  E )
2mE
h2
E 
;  is the same for both particles.
2m 2
Since, melectron  m photon implies Eelectron  E photon
Q9. Why is wave nature of matter not apparent in our
daily life?
Ans: Since
 h  W0  K .E
.
W0 copper  W0 sodium
h
2m  KE 
means,  
have
Ans: For substance with higher work function, it requires
more energy for photoelectric emission. Since work
function of copper> work function for sodium, it is difficult
to remove free electron from copper than from sodium.

h ; it means that the wavelength is too
mv
short.(You can take an example of a body of mass 1Kg
moving with a speed of 1ms-1, the wavelength comes out to
be about 6.63×10-34m)
 K .E  h
Also, 1 

h

2meV
h
2m  h 

1
hh 2


difficult
2m to
Q7. Why is it
copper than from silicon?
h
2m  KE 
h
2m

h
Q10. What is a photon? Prove that photon has zero rest
mass.
h
Ans: Photon is a packet of energy of light wave.
1
h2

remove
2m a
free
2melectron from
87
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Since m 
m0
1
2
v
c2
; where m0  rest mass of photon.
v2
c2
For photon, it travels with the speed of light. Therefore v  c
 m0  m  0  0
12.27 0
12.27 0 12.27 0
A,  / 
A
A
V
4V
2 V
/ 1



 / 

2
2

 m0  m 1 
Q11. If the intensity of incident radiation on a metal is
doubled, what happens to the K.E of electrons emitted?
Ans: Since K.E 
and intensity  no. of
photoelectrons emitted per sec, increase in intensity of
radiation will not affect K.E.
Q12. The wavelength of electromagnetic radiation is
doubled. What happen to the energy of photons?
Ans:
Energy E  h 
hc

. If the wavelength is doubled, E
will be halved.
Q13. An electron and an alpha particle have the same
de-Broglie wavelength associated with them. How are
their K.E related to each other?
UNIT VIII
ATOM AND NUCLEI
According to J.J. Thomson model, an atom is a sphere of
positive charges of uniform density of about 10 -10m
diameter in which negative charges (i.e. electrons) are
embedded like plums (i.e. fruits like dried grapes) in the
pudding.Thomson model of atom is also called ‗plum
pudding model‘. Thomson‘s model could not explain the
presence of discrete spectral lines emitted by hydrogen and
other atoms.
6.9. ALPHA-PARTICLE SCATTERING
EXPERIMENT AND RUTHERFORD‟S NUCLEAR
MODEL OF ATOM.
Ans:
Energy E  h 
hc

 mc   p 2
1
K .E  mc 2 
2
2m
2m
 K .E electron m


 K .E   particle me
2
Q14. Define threshold wavelength for photoelectric
effect?
Ans: The maximum wavelength of radiation needed to
cause photoelectric emission is known as threshold
wavelength.
Q15. De-Broglie wavelength associated with an electron
accelerated through a potential difference V is  .
What will be the wavelength when the accelerating
potential is increased to 4V?
Ans:
Experimental set-up: It consists of a narrow beam of αparticles(obtained from a radioactive substance say 214
83 Bi
88
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
kept in lead cavity) which is incident on a circular screen
coated with zinc sulphide (ZnS) after passing through a
thin gold foil (10-8m thick).when α-particles scattered by
thin gold foil fall on the screen, scintilations(i.e. short light
flashes) are produced.
OBSERVATIONS: Rutherford and his associates made
the following observations from the scattering experiment:
1.Most of the α-particles passed through the gold foil
undeflected.
2.Some of the α-particles (about 0.14% of the incident αparticles were deflected through small angles (>10).
3. A few α-particle (1 in 8000) were deflected through large
angles. Some of them even retraced their path i.e. angle of
deflection was 180o.
i.e,
1 2
1 2Ze2
mu 
2
4 0 r0
i.e,
r0 
i.e,
r0 
1
2Ze2
4 0 ( 1 mu 2 )
2
1 2Ze2
4 0 E
which is the expression for the distance of closest approach
and determines the radius or size of a nucleus.
6.11. RUTHERFORD‟S
ATOM
NUCLEAR
MODEL
OF
According to this model:
CONCLUSION: These observations led to the following
conclusions:
1. As most of the α-particles pass through the gold foil
undeflected, so it indicates that most of the space in an
atom is empty.
2. The positive charges in an atom were concentrated in a
very small region at the Centre of the atom. Rutherford
named this heavy and positively charged region as
nucleus.
3. Electrons being very light do not affect the α-particles.
6.10. DISTANCE OF CLOSEST APPROACH.
The minimum distance up to which an energetic α-particle
travelling directly towards a nucleus can move before
coming to rest and then retracing its path is known as
distance of closest approach.
Assumptions made by Rutherford.
Kinetic energy of α-particle, E=
1
2
mu
2
-------(1)
Now, electric potential energy of the system=Electric
potential at distance r0 due to nucleus × charge on αparticle
= 1
Ze
1 2Ze2
)  2e 
4 0 r0
4 0 r0 ----------(2)
(
At the distance of closest approach,
kinetic energy = potential energy.
(1) Almost all of the mass of the atom and all the positive
charges of an atom are concentrated a very small region
known as atomic nucleus.
(2) The size of the nucleus is extremely small (diameter
= 10-15m) as compared to the size of the atom (diameter
10-10m).
(3) The negatively charged particles known as electrons
revolve around but away from the nucleus. So most of the
space in an atom is empty.
(4) The number of revolving electrons is equal to the
number of positive charges in the nucleus, hence atom is
electrically neutral.
6.12. DRAWBACKS OF RUTHERFORD‟S MODEL
OF ATOM:
(1) It failed to explain the stability of the atom. Electron
revolving
around
the nucleus must lose energy
continuously. As a result of this, the radius of the path of
the electron should go on decreasing and ultimately it
should fall into the nucleus by following a spiral path. As
such, no atom should exist.
(2) It failed to explain the complex spectrum emitted by an
atom. According to Rutherford, electron can revolve
around a nucleus in circular orbits of all possible radii. So
atom can emit continuous energy spectrum. However ,
even the simplest atom i.e, hydrogen atom has line
spectrum instead of a continuous spectrum. Hence,
Rutherford‘s model of atom could not explain the
spectrum of an atom.
89
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
6.13. BOHR‟S ATOM :
To explain the stability and the radiation spectrum concept
of an atom, Niels Henrik David Bohr applied Planck‘s
quantum theory of radiation to Rutherford‘s atomic model.
He used classical as well as quantum concepts to form his
theory.
POSTULATES OF BOHR‟S ATOM MODEL:
Ln  mvrn 
nh
2
i.e, vn 
nh
2 mrn
-------- (3)
Radius of an orbit of Hydrogen atom
Substituting equation (3) in equation (2), we get
2
 nh 
e2
m
 
4 0 rn
 2 mrn 
Or n2 h2
4 mrn
2

e2
4 0
rn 
i.e,
n2 h2 0
 me2
1. Rutherford‟s model of an atom is acceptable to the extent
that an atom has a small positively charged core nucleus
where whole mass of the atom is supposed to be
concentrated.
Since,
An electron revolves around the nucleus with a definite
fixed energy in a fixed path known as stationary state
without gaining or losing the energy. The stationary state of
electron is also known as energy level.
Bohr‟s radius: The radius of the innermost orbit (n = 1 ) of
an electron in hydrogen atom is called Bohr‘s radius. It is
denoted by ao.
2. Bohr‘s postulate of quantisation of angular momentum
states that electrons can revolve only in those energy levels,
in which its angular momentum is an integral multiple of
h .
2
h 2 0
 constant, therefore, rn  n 2
 me2
We know, radius of nthorbit of hydrogen atom is given by
rn 
n2 h2 o
 me2
Here
h  6.625  10-34 Js, 0  8.854  10-12 m 2 c 2
i.e, L  mvr  nh (Bohr‟s quantization condition for
2
angular momentum)
m  9.1  1031 kg and e  1.6  1019 c; n  1
1   6.625 1034   8.854  1012 
2
 3.14  9.11031  1.6  1019 
2
2
ao 
0
Here, m = mass of an electron, v = velocity of an electron,
r = radius of the orbit, h = Planck‟s constant and n =
1,2,3…….(an integer called principle quantum number)
3. Bohr‟s postulate of early quantum concepts states that
electron can jump from higher energy level to lower energy
level radiating energy in the form of a photon.
6.14. BOHR‟S THEORY OF HYDROGEN ATOM:
Coulomb‘s force of attraction between the nucleus and the
electron revolving in an orbit of radius rn isgiven by,
Fn 
o
 5.29  1011 m  0.529 A  0.53 A
0
Thus, Bohr‘s radius = 0.53
A
Speed of an electron in an orbit of hydrogen atom,
Substituting equation (4) in equation (3), we get
vn 
nh
 me 2

2 m n 2 h 2 0
or , vn 
i.e, vn 
e2
2h 0 n
1
n
1 ee
e2
------------ (1)

2
4 0 rn
4 0 rn 2
This force provides the necessary centripetal force for the
electron to circulate

---- (2)
mvn 2
e2
e2

i.e, mvn 2 
2
rn
4 0 rn
4 0 rn
According to Bohr‘s postulate of quantization of angular
momentum,
6.15. Energy of an electron in an orbit
90
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Total energy E  KE  PE
KE 
1 2
mv
2
and PE 
1 2
1  e  e 
mv 
2
4 0
r
 E  KE  PE 

As F 
(1) When n =1 i.e, the electron is revolving in the
innermost orbit (i.e, an orbit closest to the nucleus ), then
energy of hydrogen atom is E1 = -13.6 eV. This is the
lowest energy of hydrogen atom. When the hydrogen atom
has lowest energy , the atom is said to be in the ground
state.
1  e  e 
4 0
r
1 2
1 e2
mv 
2
4 0 r
 (1)
1 e2
mv 2
and F 
(centripetal force)
4 0 r 2
r
(2) When n = 2 i.e, the electron is revolving in the second
orbit, then energy of the hydrogen atom is
mv 2
1 e2
1 e2

 mv 2 
2
r
4 0 r
4 0 r
E2 
Substituting in equation (1), we get
13.6 13.6

 3.4eV . This is the first excited
(2)2
4
state energy of hydrogen atom.
1 1 e2
1 e2
E .

2 4 0 r
4 0 r
13.6
 1.51eV . This
(3)2
E
1 e2
1 e2

8 0 r
4 0 r
(3)
E
1 e2  1 
  1
4 0 r  2 
second excited state energy of the hydrogen atom.
E
is
(4) When n  , E  13.6 /   0 .
1 e2
8 0 r
Substituting
When n  3, E3 
the
me 4
E  2 2 2
8 h 0 n
value
of
r 
n2 h2 0
 me2
RYDBERG FORMULA:
,
we
get
; using the values of mass of electron (m),
 1
1
 R 2  2
n

ni
 f
1
 where,


me 4
 1.097  107 m 1 is known as Rydberg cons tan t.
8 0 2 ch3
charge on electron (e), planck‘s constant (h), absolute
R
permittivity of free space ( 0 ), for n=1 (1st orbit)
  wavelength of spectral line emitted .
13.6
eV
n2
E1  13.6eV
The equation (1) is known as Rydberg formula for
hydrogen atom spectrum.
En  
Similarly,
13.6
eV
n2
13.6
for E2   2 eV  3.4eV (2nd excited state energy )
2
13.6
for E3   2 eV  1.51eV (3rd excited state energy )
3
13.6
for E4   2 eV  0.85eV (4th excited state energy )
4
En  
The negative sign indicates that electron and nucleus form
an attractive system. i.e, a bound system.
From the energy level diagram it is clear that when
principal quantum number (n) increases, the energy of
electron increases.
6.16. ENERGY LEVELS OF HYDROGEN ATOM
The energy of hydrogen atom in nth orbit or nth state is
given by En 
1 
 v, wave number (i.e, number of waves in unit dis tan ces)

 equation (1) can be written as,
But ,
13.6
eV
n2

 1
1 
v  R  2  2 .
n
ni 
 f
6.18. Various Spectral Series;
When electron jumps from higher energy state(orbit) to the
lower energy state (orbit) in the hydrogen atom, the
radiation of a particular wavelength or frequency (called
spectral line) is emitted.
(1) Lyman series: The spectral lines emitted due to the
transition
of
an
electron
from
any
outer
orbit(n1=2,3,4,5……….) to the first orbit (nf=1) form a
spectral series known as Lyman series. The wave number
of the spectral lines of Lyman series can be obtained from
equation (2)
91
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)

v
i.e,
1

1 1 
 R 2  2 ,
 1 ni 
where ni  2,3, 4.
Longest wavelength of the spectral line of Lyman series is
emitted when the transition of electron takes place from n i
=2 to nf = 1.
i.e,
1
min
1 1 
 R 2  2   R
1  
 1

  0 
 

This series lies in the ultra-violet region of the
electromagnetic spectrum.
(2) Balmer series : The spectral lines emitted due to the
transition of an electron from any outer orbit
(ni=3,4,5,6,………) to the second orbit (nf= 2) form a
spectral series known as Balmer series.
The wave number of the spectral lines forming Balmer
series is given by,

v
1

 1
1 
 R  2  2  , where ni  3, 4,5.......
2
n
i 

This series lies in the visible region of the electromagnetic
spectrum.
(3) Paschen Series: The spectral lines emitted due to the
transition of an electron from any outer orbit (ni =
4,5,6……) to the third orbit (nf = 3) form a spectral series
known as Paschen series.
The wave number of the spectral lines forming Paschen
series is given by,
1
1 
u   R 2  2 

3
n
i 

1
where ni  4,5, 6......
This series lies in the Infra-red region of the
electromagnetic spectrum.
(4) Brackett Series: The spectral lines emitted due to the
transition of an electron from any outer orbit
(ni = 5,6,7……) to the fourth orbit (nf =4) form a spectral
series known as Brackett series.
(5) Pfund Series: The spectral lines emitted due to the
transition of an electron from any outer orbit (ni =
6,7,8……) to the fifth orbit (nf = 5 ) form a spectral series
known as Pfund series .
The wave number of the spectral lines forming Pfund series
is given by,

u
1

1 1 
 R  2  2  , where ni  6,7,8........
 5 ni 
This series lies in the far Infra- red region of the
electromagnetic spectrum.
Drawbacks of Bohr‟s Atomic Model:
Bohr‘s atomic model has the following limitations:
1. It successfully explains the spectra of simple atoms ( i.e,
the atoms having only one electron). For example, it can
explain the spectra of hydrogen atom and hydrogen like
atoms (He+, Li++etc). But this model could not explain the
spectra of complex atoms having more than one electron.
2. When the spectral lines of a Balmer series was observed
under a powerful microscope, it was found that each
spectral line consists of closely spaced lines. Bohr‘s model
of atom could not explain this fine structure of the spectral
lines of Balmer series.
3. Bohr‘s atomic model does not give any indication
regarding the arrangement and distribution of electrons in
an atom.
4. This model could not account for the wave nature of
electrons.
6.19. COMPOSITION OF NUCLEUS AND ATOMIC
MASS
Atom is very small and its nucleus is very very small. The
mass of the nucleus and its constituent particles is
expressed by a very small unit called unified mass unit (u).
This unit is also known as atomic mass unit (a.m.u).
The wave number of the spectral lines forming Brackett
series are given by ,

u
1

1
1 
 R 2  2 
4
n
i 

where ni  5,6,7.......
This series lies in the far Infra- red region of the
electromagnetic spectrum.
92
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Atomic mass unit is defined as
1
th of the mass of carbon -12
12
number (i.e, N=6.0225 10 23 )
Thus, 1 mole of carbon atom (i.e, 12g )
Q. Is the neutron outside nucleus stable?
Ans. No.
Q. What are thermal neutrons?
Ans. The neutrons are then in thermal equilibrium
with the molecules of the moderator. Such
neutrons are called thermal neutrons. The speed of
thermal neutrons is 2200 ms-1 and energy = 0.025
contains 6.0225  10 23 atoms.
eV.
atom (i.e, 6 C12 ). Mass of a 6 C12
(carbon atom) is nearly 1.993  10-26 kg.
According to Avogadro's hypothesis, 1 mole ( 6 C12 ) of a
substance contains atoms equal to Avogadro's
 mass of 6.0225 10 23 carbon ( 6 C12 ) atoms 12 g
or, mass of 1 atom of carbon 6 C12 =
12g
6.0225 1023
According to definition,
1
1 a.m.u (or 1u) =  mass of 1 atom of carbon ( 6 C12 )
12
1
12g
= 
 1.6604 10 -24 g
12 6.0225 10 23
or,
1 a.m.u (or 1 u) =1.66 10-27 kg.
Q. What are the constituents of a nucleus?
Ans.A nucleus is composed of protons and neutrons. The
sum of the number of protons (Z) and the number of
neutrons (N) is called mass number (A).
A=Z+N
where A= mass number
N= no. of neutrons
Z= atomic no.
Properties of protons and neutrons.
Positive charge of an atom is concentrated in the nucleus
due to positively charged particles known as Protons.
1. Proton is a constituent of nucleus.
2. Proton is a positively charged particle.
3. The magnitude of charge on a proton = 1.6 × 10-19 C.
4. Its mass is equal to 1.6726 × 10-27 kg.
5. It has spin (intrinsic angular momentum) equal to 1/2.
6. The number of protons in the nucleus is called atomic
number (Z) of the atom.
Properties of Neutron
1. Neutron is a constituent of the nucleus.
2. Neutron is a neutral particle i.e. it has no charge on it.
3. The mass of neutron is about 1.6748 x 10 -27 kg.
4.
Neutron has very high penetrating power. Being
neutral, it is neither attracted nor repelled by the nucleus of
an atom, so it can penetrate deep into the atom.
5. Neutron has low ionizing power.
6. Neutron inside the nucleus is stable.
6.20.Isotopes Isobars ans Isotones
Isotopes : The atoms of an element having same atomic
number (Z) but different mass number (A) are called
isotopes. 1H1, 1H3
(ii) Isobars :The atoms of elements having same mass
number (A) but different atomic number (Z) are called
Isobars.
Isobars have same number of total nucleons but different
number of protons, electrons and neutrons.
For example :1H3 and 2He3; 3Li7 and 4Be7; 7N15 and
15
8O ,
(iii) Isotones : The atoms of the elements whose nuclei have
the same number of neutrons are called Isotones.
For example :4Be9 and 5Bl06C13 and 7N14

Mass number is the number of protons and
neutrons in the nucleus and hence always an
integer.
6.0.Mass Defect
Mass Defect is defined as the difference between the mass
of the constituent nucleons of the nucleus in the free state
and the mass of the nucleus.
Consider a nucleus of mass M having Z protons and (A - Z)
neutrons. Let mnbe the mass o each neutron and mpbe the
mass of each proton.
Mass of Z proton = mpZ and mass of (A-Z)
neutrons= mn(A-Z)
Sum of masses of constituent nucleons = mpZ+ mn(A-Z)
Hence mass defect
M=mass of nucleus.
={mpZ+ mn(A-Z)}-M ; where
Units of Mass Defect :(i) Mass defect is measured in a.m.u.
if atomic masses are expressed in a.m.u.
(ii) If atomic masses are expressed in kg, then mass defect
is also measured in kg.
93
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Number of neutrons greater than the number of
protons for heavy nuclei
As mass number (A) increases i.e., number of protons
increases, then the electrostatic force of repulsion between
protons increases. As a result of this, the nucleus may not
exist. Since a nucleus having large number of protons exists
(e.g. 92U235 contains 92 protons), so the electrostatic force
of repulsion between protons is to be balanced for the
existence of the nucleus. To compensate the repulsive
force, number of neutrons in the nucleus increases. Thus, a
larger or heavy nucleus has more neutrons than protons.
2. beta-particle has a negative charge equal to the charge
on an electron.
3. The rest mass of beta-particle is equal to the mass of an
electron.
4. They are emitted with a velocity of the order of 2.97 ×
108ms-1.
5. They are deflected by electric and magnetic fields.
6. They cause fluorescence in zinc sulphide and other
fluorescent materials.
7. The penetration power of particles is more than that of
a-particles.
8. They affect the photographic plates.
Properties of
Rutherford‟s experiment performed to study the
Becquerel rays.
Ans. A small sample of radioactive element (radium) is
placed in a hole made in a block of lead. A narrow beam of
rays emerges from the top of the block. When oppositely
charged plates are placed at the sides of this beam of rays
some of the rays deflected to the left, some to the right and
some are not deflected at all. A magnetic field around the
beam of rays shows the same effect. The rays deflected to
the left are positively charged and known as beta–rays
or beta-particles.
The rays deflected to the right are negatively charged and
known as rays or alpha-particles. The rays which are not
deflected are uncharged and known as gamma-rays or
photons.
Antineutrino
Antineutrino is an anti particle of neutrino.
The mass of neutrino is zero. The charge on neutrino is
zero i.e. it is a neutral particle. Neutrino and antineutrino
are emitted to conserve energy and momentum during
decay process.
Properties of alpha, beta and gamma rays
Properties of α particle:
l. An α-particle is equivalent to a helium nucleus (2He4)
consisting of two protons and two neutrons.
Properties of beta-particle.
1. beta-particle is a fast moving electron (-1e°).
ray
1. They are the packets of energy of electromagnetic
radiations.
2. They have no charge.
3. The rest mass of gamma-rays is zero.
4. They always travel with the speed of light in vacuum (3
x 108ms-1).
5. They are not deflected by electric and magnetic fields.
6. They cause ionisation but their ionising power is about
times the ionising power of beta particles.
7.
They have very high penetration power. Their
penetration power is 100 times greater than beta particles.
8. They affect the photographic plates more than betaparticles.
6.1.Radioactivity
The phenomenon of spontaneous emission of radiations by
heavy elements is called radio-activity.
Laws of radioactive decay.
Ans. Radioactive decay obeys the following laws :
1. Radioactive decay (i.e. disintegration) is a spontaneous
process and is not affected by the external conditions such
as temperature, pressure etc,
2. When a radioactive element decays by emitting an aparticle, its position goes down by two places in the
periodic table.
Z
2. They have positive charge equal to + 2e, where
e = 1.6×10-19 C.
3. They are emitted with velocities ranging between
1.4 × 107ms-1 to 2.2×107ms-1.
4. They are deflected by electric and magnetic fields.
S. They cause fluorescence in certain materials like barium
platinocyanide, zinc sulphide etc.
6. They have low penetrating power.
7. They have rest mass equal to four times the mass of a
proton
8. They have high ionising power.
9. They slightly affect the photographic plates.

  particle
X A 
Z  2 Y A 4
The original radioactive element (zXA) is called parent
element and the product element obtained after alpha decay is called daughter element.
3.
When a radioactive element decays by emitting a
particle, its position is raised by one place in the periodic
table.
Z
  particle
X A 
Z 1 Y A
4. When a radioactive element decays by emitting a -rays,
its position remains the same in the periodic table. The
radioactive element in the excited state comes to its ground
state by emitting the energy in the form of a photon or yray.
Z
X
*
A
  particle
(excited state) 
Z Y A ( ground state)
94
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
5. The rate of disintegration of a radioactive substance is
directly proportional to the number of atoms remained
undecayed in the substance. This law is called radioactive
decay law or disintegrationlaw.
corresponding to Iron (26Fe56) nucleus. Thus, iron is a
stable element.
For nuclei having A> 56, binding energy per nucleon
gradually decreases. For uranium (A =238), one of the
heaviest natural element, the value of B.E/nucleon drops to
7.5 MeV.
Binding energy
The energy required to disintegrate a nucleus completely is
known as binding energy.
Binding energy = mc2.
or B.E. = {{mpZ + mn(A - Z)} - M] c2
When mass defect (Am) is expressed in a.m.u., then the
binding energy is written as
B.E. = [{mpZ + mn(A~ Z)} - M] × 931 MeV
or B.E. = m x 931MeV
6.2. Binding energy pernucleon(B.E/nucleon)
The average energy required to release a nucleon from the
nucleus is called binding energy per nucleon.
BE/nucleon =
m931MeV
A
Importance of BE/nucleon
Binding energy per nucleon determines the stability of a
nucleus. That is, stability of a nucleus is proportional to the
binding energy/ nucleon.
If binding energy per nucleon of a nucleus is less, the
nucleus is less stable whereas the nucleus is more stable if
its binding energy per nucleon is higher.
(i) Some nuclei with mass number A< 20 have large
binding energy per nucleon than their neighbouring nuclei.
4
8
12
16
20
2He , 4Be , 6C , 86 and 10Ne have more binding
energy per nucleon than their neighbours.
So these nuclei are more stable than their neighbours
Conclusions :
(i) The binding energy per nucleon has low value for both
the light and heavy nuclei. So they are unstable nuclei.
(ii) The intermediate nuclei have large value of binding
energy per nucleon, so they are more stable.
6.3. Nuclear reaction
The process by which the identity of a nucleus is changed
when it is bombarded by an energetic particle is called
nuclear reaction.
Q-value and significance
Q-value or Energy of Nuclear Reaction. The energy
absorbed or released during nuclear reaction is known as
Q-value of Nuclear reaction.
Q-value is defined as the difference between the mass of
reactants
and
the
mass
of
products
i.e.
2
Q value = (Mass of Reactants - Mass of products) c joule
or Q = (Mass of reactants - Mass of products) a.m.u.
If Q < 0, the nuclear reaction is known as endothermic.
The energy is absorbed in the reaction.
If Q > 0, the nuclear reaction is known as exothermic. The
energy is released in the reaction.
Physical quantities conserved during a nuclear reaction
(i) Total charge or atomic number (Z) before and after the
reaction is conserved.
4
14
17
1
2He + 7N =>8O + 1H
Z = 9 before the reaction and Z = 9 after the reaction.
(ii) Mass number (A) is conserved.
A = 4 + 14 = 18 before the reaction and A = 17 + 1 = 18
after the reaction
(iii) Linear momentum of the particles before the reaction
is equal to the linear momentum of the particles after the
reaction. That is linear momentum is conserved.
(v)
Angular momentum of the particles before
the reaction is equal to the angular
momentum of the particles after the
reaction. That is angular momentum is also
conserved.
Nuclear fission reaction
Nuclear fission is a process of splitting a heavy nucleus
into two nuclei of comparable masses along with the
emission of large amount of energy.
For A > 40, binding energy per nucleon increases gradually
till it attains a maximum value 8.8 MeV per nucleon
Theory of nuclear reaction (liquid drop mode)
95
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Theory of Nuclear Fission The nuclear fission was
explained by Bohr and Wheeler using liquid drop model of
the nucleus.
According to liquid drop model, the nucleus is a drop of
incompressible electrically charged nuclear liquid. There
are two forces in the nucleus i.e. surface tension like
nuclear force and the coulomb‟s repulsive force. Nuclear
force tends to keep the spherical shape of the nucleus. On
the other hand, Coulomb‘s repulsive force tends to destroy
the spherical shape of the nucleus. The shape of the nucleus
remains spherical as long as nuclear force balances the
Coulomb‘s repulsive force.
When the nucleus is bombarded with a neutron, the neutron
is captured by the nucleus. This provides an excitation
energy to the nucleus. The excitation energy favours the
Coulomb‘s repulsive force and tends to destroy the
spherical shape of the nucleus. As a result of this,
oscillations are set up within the nucleus whose shape goes
on deforming. Ultimately, the shape of the nucleus
becomes like a dumb-bell. If the excitation energy is large
enough, the Coulomb‘s repulsive force pushes the two balls
apart, thereby splitting the nucleus into two nuclei of
comparable masses.
6.5. Chain reaction and reproduction factor
(k)
In nuclear fission, three neutrons are produced along with
the release of large amount of energy. These newly
produced neutrons can cause further fission of more nuclei,
producing large number of neutrons. The process continues
and known as a chain reaction.
Un-controlled chain reaction : If more than one neutrons
produced in a fission cause further fissions at each stage,
then the number of fissions and energy released multiply
rapidly. Such a chain reaction is called un-controlled
chain reaction. In such a chain reaction, huge amount of
energy is released within a fraction of a second. This is the
underlying
principle
of
atom
bomb.
Radiative capture
If the excitation energy is small, the nucleus simply
emits -rays of energy equal to the excitation energy
and the „surface tension like nuclear force‟ restores
the spherical shape of the nucleus. This process is
called a radiative capture than the fission.
6.4. Energy released in per fission of U235
Nuclear fission of 92U235 is represented as follows :
1
235
141 + Kr92+ 3 n1 + Q (Energy)
0n + 92U ——>56Ba
36
0
The energy released during the fission of 92U235 is
equivalent to the difference between the mass of reactants
and the mass of products.
Mass of 92U235 = 235.124 a.m.u.
Mass of 0n1 = 1.009 a.m.u.
Mass of reactants (92U235 and 0n1) = 236.133 a.m.u.
Mass of 56Ba141 = 140.958 a.m.u.
Mass of36Kr92 = 91.926 a.m.u.
Mass of 3 0n1 = 3.027 a.m.u.
Mass of products = 235.911 a.m.u.
Mass of defect = mass of reactants - Mass of products
= 236.133 -235.911 = 0.222 a.m.u.
Since1 a.m.u. = 931 MeV
Energy released per fission of 92U235 = 0.222 ×931
= 206.682 MeV 200 MeV
Controlled chain reaction : If only one neutron is
available to cause further fission at each stage, then a
constant amount of energy is released. Such a reaction is
called controlled chain reaction.
Reproduction factor (k): It is the ratio of rate of
production of neutron to rate of loss of neutron.
If k = 1 i.e. the rate of production of neutron is equal to the
rate of loss of neutron, the mass of the fissionable material
is said to be critical and the chain reaction is sustained.
If k <1, then chain reaction stops and if k>1, chain
reaction is accelerated.
Sustained nuclear reaction
Another method to sustain the chain reaction is, to slow
down the neutrons emitted in the fission. The chances of
causing fission of 92U238 by slow neutrons are very small
and hence these neutrons are available for the fission of
235 isotope. The fast neutrons can be slowed down by
92U
certain materials called moderators. Graphite and heavy
water are the examples of moderators.
96
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Breeder Reactors. These reactors are used for
breeding of nuclear fuel.
6.6. Nuclear Reactor
achieved. The cadmium rods are used to control the chain
reaction. The fission produces heat in the nuclear reactor
core. The coolant transfers this heat from the core to the
heat exchanger, where steam is formed. This steam
produced at a very high pressure runs a turbine and the
electricity is obtained at the generator.
6.7. Nuclear fusion and its origin
A process in which two very light nuclei (A < 8) combine to
form a nucleus with a larger mass number along with
simultaneous release of large amount of energy is called
nuclear fusion.
For example, (i) when two nuclei of deuterium 1H2) fuse
together, the following products are formed :
Nuclear reactor is a device in which nuclear fission is
maintained as a self-supporting yet controlled chain
reaction. It was formerly known as atomic pile.
1. Fissionable material (Fuel) : The fissionable material
used in the reactor is called the fuel of the reactor. Uranium
isotope (U235), Thorium isotope (Th232) and Plutonium
isotopes (Pu239, Pu240 and Pu241) are the most
commonly used fuels in the reactor.
2. Moderator : Moderator is used to slow down the fast
moving neutrons. Most commonly used moderators are
graphite and heavy water.
3. Control Material: Control material is used to control
the chain reaction and to maintain a stable rate of reaction.
This material controls the number of neutrons available for
the fission. For example, cadmium rods are inserted into
the core of the reactor because they can absorb the
neutrons. The neutrons available for fission are controlled
by moving the cadmium rods in or out of the core of the
reactor.
4.Coolant : Coolant is a cooling material which removes
the heat generated due to fission in the reactor. Commonly
used coolants are water, CO2 , nitrogen etc.
5.Protective Shield : A protective shield in the form a
concrete thick wall surrounds the core of the reactor to save
the persons working around the reactor from the hazardous
radiations.
Working :A few 92U235 nuclei undergo fission liberating
fast neutrons. These fast neutrons are slowed down to an
energy about 0.025 eV by the surrounding moderator
(graphite) through elastic collisions. When the reactor
becomes critical, self sustained controlled chain reaction is
1H2 + 1H2——>2He3 + onl + 3.3 MeV
Binding energy per nucleon of very light nuclei is less than
that of intermediate nuclei. It means light nuclei are less
stable than that of intermediate nuclei. This shows that the
sum of masses of the individual nuclei is more than the
mass of the nucleus formed by their fusion. The difference
in mass (called mass defect) is released in the form of
energy.
Nuclear fusion reaction is called a thermonuclear
reaction
Nuclear fusion cannot take place so easily. When two light
nuclei are brought closer to each other, they exert a
repulsive force on each other due to their positive charges.
As such, these nuclei cannot fuse together. These nuclei
can fuse together if they have enough kinetic energy to
overcome the force of repulsion between them. High
kinetic energy implies a high Temperature. It means that a
huge temperature is required for the process. Hence it is
rightly called a thermonuclear reaction.
Uncontrolled fusion reaction and its application
To achieve the very high temperature (107K) to start a
nuclear fusion reaction, an atom bomb employing the
process of nuclear fission is used. Thus, atomic explosion
triggers the fusion process and simultaneous release of
tremendous amount of energy. This is known as
uncontrolled fusion reactions and is the principle
ofhydrogen bomb.
Radiation hazards and useful applications of nuclear
radiations
.(i) Radiation damage to the chromosomes in the
reproductive organs can cause genetic disorder.
(ii) Radiation damage to the blood producing cells in the
spleen can increase the possibility of contracting leukemia.
(iii) An acute exposure to radiation weakens or even
destroys the body immune system and may lead to
death.
(iv) Long exposure to radiations causes cancer.
(v) Long exposure to radiations causes blindness.
(vi) Besides external exposure, radiation damage can come
from inhaling air containing radio-isotopes and eating food
97
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
contaminated with radio-isotopes.
ADDITIONAL QUESTIONS
Q.
A neutron strikes a 5B10 nucleus with the
subsequent emission of an a-particle. Write the
corresponding nuclear reaction. Find the atomic
number, mass number and the chemical name of the
resulting nucleus.
Ans. 0n1 + 5B10——>ZYA + 2He4
(alpha-particle).
Ans: According to the conservation of atomic number
0+5 = Z + 2 so Z = 5 - 2 = 3.
According to the conservation of mass number 1 + 10 = A
+ 4 So A = 11 -4 = 7.
A
7
7
7
ZY = 3Y i.e. 3Li . So the resulting nucleus is 3L1 .
Hence the nuclear reaction is:
1
10
7
4
0n + 5B —>3Li + 2He .
Q. Why a nucleus can eject electrons (betaparticles) though it contains no electrons?
Or
How are beta-rays emitted from a nucleus, when it does
not contain electrons ?
Ans.In fact, during the decay of a nucleus, a neutron is
converted into a proton, an electron and antineutrino.
i.e. 0nl———>1H1 + -1e° (beta particle) + antineutrino.
The mass of neutron is greater than that of a proton. A part
of this mass is used up to form an electron and the
remaining part is converted into energy which is shared by
emitted electron and the antineutrino.
Q. Activity of a radioactive substance having short
half-life is much more than a substance having
long half-life. Explain why ?
Ans. Activity of a radioactive substance is the rate of
disintegration of the substance and is inversely proportional
to its half-life
When half life (T) is small, activity is large and vice-versa.
However, the controlled exposure to radiations has
number of uses :
1. The dangerous disease like cancer is cured by radiation
therapy, gamma rays from Co60 are used for this purpose.
2. Radiation are used to induce plant mutations which
improves the varieties of many crops such as wheat, peas
and rice.
3. Radiations are also used to eliminate agriculture pests.
4. Radiations like X-rays are used to detect the fracture in
the bone and presence of foreign material in the human
body.
5. Gamma rays or X-rays are used to detect the defects in
metal castings and welds.
Q. Why nuclei have mass less than the sum of the
masses of the individual nucleons in them ?
Ans. When nucleons are brought to form a nucleus, a
fraction of their mass is converted into the energy to hold
these nucleons in the nucleus. Hence, mass of a nucleus is
less than the sum of the individual nucleons in them.
Q. Why is a neutron preferred as a bombarding
particle ?
Ans. Neutron is a neutral particle i.e. it has no charge. So it
is neither repelled nor attracted by the nucleus and hence
can penetrate deep into the nucleus to cause a nuclear
reaction.
Q. Explain why heavy nuclei are unstable?
Ans, The binding energy per nucleon of a heavy nucleus is
small. It means, small energy is required to extract a
nucleon from a heavy nucleus. That is why heavy nuclei
are unstable.
98
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Q. In heavy nuclei, the number of neutrons is more than
number of protons. Explain why ?
Ans. In a nucleus, there is a Coulomb‘s repulsive force
between protons in addition to a strong attractive nuclear
force. On the other hand, in case of neutrons with in the
nucleus, there is only a short range attractive nuclear force.
For heavy nuclei to be stable, the repulsive force must be
less. This is possible only if the number of the neutrons is
more than the number of protons.
Q. Explain
why 92U238 is not suitable for chain
reaction ?
Ans. The energy required to disintegrate 92U238 nucleus is
larger than the energy required to disintegrate92U235
nucleus. A neutron of energy 1.2 MeV is required to
disintegrate 92U238 nucleus. Since neutrons of this much
energy are rarely available and hence 92U238 is not suitable
for chain reaction.
Q. It is very difficult to initiate the nuclear fusion.
Explain the statement.
Ans. Nuclear fusion can be carried out at extremely high
temperature (=107K). Since this much temperature cannot
be generated in any furnace, so nuclear fusion cannot be
initiated. Moreover, if this amount of temperature is
generated by atomic explosion, then it is difficult to contain
the material used in fission. Hence, nuclear fusion cannot
be carried out easily.
Q. Is the mass number an integer or odd?
Ans. Mass number is the number of protons and neutrons in
the nucleus and hence always an integer.
Ans. According to Einstein, mass and energy are interconvertible. The Einstein‘s mass-energy relationship is
given by E = mc2
where E is energy, m is mass and c is velocity of light in
vacuum. It means that the mass of
a body can be
represented in terms of energy and vice-versa.
Q. What are hydrogen like atoms?
Ans. Hydrogen-like atom consists of a nucleus having
charge + Ze (where Z is the atomic number of the atom)
and an electron of charge -e revolves round the nucleus in
circular orbit. For example, singly ionized helium atom
He+ and doubly ionized lithium atom Li++ are hydrogen
like atoms.
UNIT IX
ELECTRONIC DEVICES
Valence band: The energy band occupied by the valence
electrons is known as valence band.
Conduction band: The energy band which is partially
filled or empty that lies above the valence band is called as
conduction band.
Forbidden energy gap: The gap between the conduction
band and the valence band is known as the forbidden
energy gap.
Energy band in a solid
Q. Find the size of a nucleus in terms of mass number.
Ans.
Volume of nucleus  Mass number
So,
4
 R3  A
3
1
 R  R0 A 3 ; where R0  Rydberg ' s cons tan t
 1.2 10 15 m
The range of energies possessed by an electron in an atom
is known as energy band in solids.
Q. What is radioactive decay constant? Write its units.
Ans. Radioactive decay constant  is the reciprocal of the
time during which the number of atoms in the radioactive
substance reduces to 36.8% of the original number of
atoms in it.
Units of decay constant. Decay constant is expressed in s1 or min-1or day-1or year-1.
6.8. Classification of solids as insulators,
conductors and insulators using the idea
of energy band theory
(i) Insulators: Insulators (e.g. wood, glass, etc.) are those
substances which do not allow
the passage of electric current through them. In terms of
energy band, the valence band
Q. Explain Einstein‟s mass-energy formula.
99
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
is
full
while
the
conduction
band
is
empty.
Further, the energy gap between valence and conduction
bands is very large (~ 15 eV)
as shown in Fig. Therefore, a very high electric field is
required to push the valence electrons to the conduction
band.
(ii). Conductors:
Conductors (e.g. copper, aluminium) are
those substances which easily allow the
passage of electric current through them. It
is becausethere are a large number of free
electrons available in a conductor. In terms
of energy band, the valence and conduction
bands overlap each other as shown in Fig,.
Due to this overlapping, a slight potential
difference acrossa conductor causes the free
electrons to constitute electiic current.
(iii). Semiconductors:
Semiconductors (e.g. germanium, silicon,
etc.) are those substances whose electrical
conductivity lies in between conductors and
insulators. In terms of energy band, the
valence band is almost filled and conduction
band is almost empty. Further, the energy
gap between valence and conduction bands
is very small as shown in Fig. Therefore,
comparatively smaller electric field (smaller
than insulators but much greater than
conductors) is required to push the electrons
from the valence band to the conduction
band. Semiconductor has small energy gap
(= leV) between valence and conduction
bands.
Doping :The process of adding impurities in the intrinsic
semiconductor is called doping. Methods of Doping
(i) A very small quantity of impurity atoms is made to
diffuse into the high purity molten
material such as germanium when the crystal is grown out
of melt.
(ii) Impurity atoms can also be added into the intrinsic
semiconductor by heating it in the environment having
impurity atoms.
(i) Intrinsic semiconductor: A semiconductor in its pure
form is known as intrinsi semiconductor.
(ii) Extrinsic semiconductor: A semiconductor in its
impure form obtained by adding impurity to a pure
semiconductor is known as extrinsic semiconductor.
Pentavalent impurity :The elements whose each atom has
five valence electrons are called pcntavalent impurities.
For examples, Arsenic (As), Antimony (Sb), Phosphorous
(P) etc.
These impurities are also known as donor impurities as
they donate extra free electrons to the semiconductor.
Trivalent impurities ;The elements whose each atom has
three valence electrons are called trivalent impurities. For
examples, Indium (In), Gallium (Ga), Aluminium (Al),
Boron (B) etc.
These impurities are also known as acceptor impurities as
they accept electrons from the covalent bonds of the
semiconductor.
6.9. P-type
or
p-type
Semiconductor.
When trivalent impurity is added to pure
germanium or silicon crystal, we get extrinsic
semiconductor known as P-type semiconductor.
Trivalent impurity atom say indium has three valence
electrons. When an atom of indium is added into the silicon
crystal, this atom replaces one of the Silicon atom and
settles in the lattice site of replaced silicon atom.
This indium atom forms three covalent bonds with the
neighbouring three silicon atoms. The fourth bondis
incomplete which has a deficiency of one electron. This is
called hole and behaves like a positively charged
particle. This hole attracts the electron from the
neighbouring covalent bond to fill itself. Now a new hole is
created at the site from which the electron has been
attracted to fill the hole. A large number of holes are
formed by adding more and more trivalent (indium) atoms
in the silicon crystal.
The crystal of this type has an excess of holes or positive
charge carriers and hence known as P-type semiconductor
or positive type semiconductor. Majority charge carriers in
p-type semiconductor are holes and the minority carriers
are electrons .
6.10.
N-type Semiconductor.
100
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
When pentavalent impurity atoms are added
to pure germanium or silicon crystal, we get an extrinsic
semiconductor known as N-type semiconductor.
carriers are holes.
When arsenic atom is added into the silicon crystal, it
replaces one of the silicon atom and settles in the lattice site
of replaced silicon atom. This arsenic atom forms four
covalent bonds by sharing its four electrons with the
neighbouring silicon atoms. The fifth valence electron of
arsenic remains un-accomodated. Now it becomes a free
electron and wander through the crystal . In this way a large
number of free electrons are available in the crystal when a
small amount of arsenic is added to the silicon crystal.
The crystal of this type has excess number of electrons or
negatively charged carriers and hence it is known as Ntype or negative type semiconductor.
When a semiconductor crystal (germanium or silicon) is so
prepared that one half is p-type and the other n-type, the
contact surface dividing the two halves is called a p-n
junction.
Barrier Potential
The diffusion of electrons and holes establishes a potential
difference across the junction. This is called potential
barrier or junction barrier V0
6.11.

Majority
charge
carriers
in
N-type
semiconductor are electrons and the minority
charge carriers are holes which are thermally
generated.
INTRINSIC
SEMICONDUCTORS
1.
Intrinsic
semiconductors are the
crystals of pure
elements like germanium
and silicon.
2.
In
intrinsic
semiconductor, the number
density of
electrons is equal to the
number density of holes.
i.e., ne = nh.
3.
The
electrical
conductivity of intrinsic
semiconductors is low.
4.
The
electrical
conductivity of intrinsic
semiconductors
mainly depend on their
temperatures.
EXTRINSIC
SEMICONDUCTORS
1.When some impurity is
added in the intrinsic
semiconductor, we get
an
extrinsic
semiconductors.
2.
In
extrinsic
semiconductor,
the
number density of
electrons is not equal to the
number density of holes.
i.e.,
3.
The
electrical
conductivity of extrinsic
semiconductors
is high.
4.
The
electrical
conductivity of extrinsic
semiconductors
depend on the temperature
as well as the amount of
impurity added in them
Difference
between
semiconductors
n-type
and
p-type
n-type
1.
When pentavalent
impurity atoms like As, Sb
etc. are
added in the intrinsic
semiconductor, we get ntype
semiconductor.
2. The majority carriers in
n-type semiconductor are
electrons and minority
p-type
1. When trivalent impurity
atoms like gallium, indium
etc.
are added in the
intrinsic semiconductor, we
get
p-type semiconductor.
2. The majority carriers in
p-type semiconductor are
holes and minority carriers
are electrons.
p-n junction
Depletion layer: The region near the junction consisting of
immobile positive and negative ions. Its thickness is about
10-3cm.
Forward biasing: When external voltage applied to the
junction is in such a direction that it cancels the potential
barrier, thus perrnuing current flow, it is called forward
biasing.
(i) The potential barrier is reduced and at some forward
voltage (0.1 to 0.3 V), it is eliminated altogether.
(ii) The junction offers low resistance (called forward
resistance Rf) to current flow.
With forward bias to the pn junction i.e. p-type connected
to positive terminal and n-type connected to negative
terminal, the potential barrier is reduced. At some forward
voltage (0.7 V for Si and 0.3 V for Ge), the potential barrier
is altogether eliminated and current starts flowing in the
circuit. From now onwards, the current increases with the
increase in forward voltage. Thus a rising curve OB is
obtained with forward bias as shown in Fig.
Reverse biasing: When the external voltage applied to
the junction is in such a direction that potential barrier
increases, it is called reverse biasing. During reverse
biasing: (i) The potential barrier is increased.
(ii)The junction offers very high resistance (called reverse
resistance Rr) to current flow.
(iii)
No current flows in the circuit due to the
establishment of high resistance path.
With reverse bias to the pn junction i.e. p-type connected to
negative terminal and n-type connected to positive
terminal, potential barrier at the junction is increased.
Therefore, the junction resistance becomes very high and
101
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
practically no current flows through the circuit. However,
in practice, a very small current (of the order of micrompere) flows in the circuit with reverse bias as shown in
the reverse characteristic.
Breakdown voltage: It is the reverse voltage at which pn
junction breaks down with sudden rise in reverse current.
(ii) Knee voltage: It is the forward voltage at which the
current through the junction starts to increase rapidly.
D.C. or Static resistance of the diode is defined as the
ratio of the d.c. voltage across the diode to the direct
current flowing through it.
Rdc 
V
I
A.C. or Dynamic resistance of the diode is defined as the
ratio of the small change in voltage to the corresponding
small change in current in the diode.
Rac 
V
I
6.12.
Rectifier
A device which converts alternating current (a.c.) into
direct current (d.c.) is called rectifier. The process of
convening a.c. into d.c. is called rectification.
Principle :Junction diode conducts only when forward
biased and it does not conduct when reverse biased. This
fact makes the junction diode to work as a rectifier.
1.
Q. Why filter circuits are used in a full wave
rectifier?
Ans. The output is again fluctuating (or pulsating
d.c.) which can be smoothened by using a filter
circuit
Q. What is a zener diode?
Ans. Zener diode is a properly doped diode that can
work even in the breakdown region.
The direction of induced e.m.f. is such that the upper end of
the secondary coil becomes positive and lower end
becomes negative. Since upper end of secondary coil is
connected to p-region and lower end is connected to the nregion of the junction diode, so the junction diode is
forward biased during the positive half of input a.c. Thus
the junction diode conducts. The output voltage is
obtained across the load
resistance RL.
When negative half cycle of a.c. input flows through the
primary coil, again induced e.m.f. is set up across the
secondary coil due to mutual induction. Now the direction
of induced e.m.f. is such that upper end of the secondary
coil becomes negative and lower end becomes positive. So
the junction diode is reverse biased. Hence the junction
diode does not conduct and no output across the load
resistance during negative half of input a.c.
Disadvantages :
1. Since the output signal is discontinuous, so the
efficiency of half wave rectifier is small.
2. The output is not pure d.c. but it is a fluctuating (or
pulsating d.c.) which contains a.c. components or ripples
also.
Half wave rectifier
When the positive half of a.c. input signal flows through
the primary coil, an induced e.m.f. is set up in the
secondary coil due to mutual induction.
2. Full wave rectifier
Full wave rectifier rectifies both halves of a.c. input signal.
The a.c. input signal is fed to the primary (P) coil of the
transformer. The p-regions of both the diodes D1 and D2
are connected to the two ends of the secondary coil (S).
Working : When positive half cycle of input a.c. signal
flows through the primary coil, induced e.m.f. is set up in
the secondary coil due to mutual induction. The direction of
induced e.m.f. is such that the upper end of the secondary
coil becomes positive while the lower end becomes
negative. Thus, diode D1is forward biased and diode
102
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
D2
is
reverse
biased,
So
the
current
due
to
diode D1flows through the circuit in a direction shown by
arrows (above RL). The output voltage is obtained across
the load resistance (RL).
During negative half cycle of input a.c. signal, diode D1 is
reverse biased and diode D2 is forward biased. The current
due to diode D2 flows through the circuit in a direction
shown by arrows (below RL). The output voltage is
obtained across the load resistance (RL).
Types: (i) n-p-n transistor
When a thin layer of p-type semiconductor is sand witched
between two thick layers of n-type semiconductor, we get a
transistor which is known as n-p-n transistor.
(ii) p-n-p transistor.
Advantage: In full wave rectifier, output is continuous, so
its efficiency is more than that of the half wave rectifier.
When a thin layer of n-type semiconductor is sand witched
between two thick layers of p-type semiconductors, we get a
transistor which is known as p-n-p transistor.
Note: E=emitter, C=collector and B=base.
Action of n-p-n transistor
INPUT AND OUTPUT WAVES OF halve wave
rectifier (above) and full wave rectifier (below)
6.13.
Transistor
A transistor is a three terminal semiconductor formed when
a thin layer of one type of p-type or n-type is sandwitched
between two thick layers of other type of semiconductor.
The emitter-base junction of n-p-n transistor is forward
biased whereas the collector base junction is reversed
biased.
When emitter-base junction is forward biased, electrons
(majority carriers) in the emitter are repelled towards base.
103
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
The barrier potential of emitter-base junction decreases and
the electrons enter the base. About 5% of these electrons
combine with the holes in the base region resulting in small
base current (Ib). The remaining electrons (=95%) enter the
collector region because they are attracted towards the
positive terminal of the battery connected with basecollector junction. For each electron entering the positive
terminal of the battery connected with collector-base
junction, an electron from the negative terminal of the
battery connected with emitter-base junction enters the
emitter region. Thus continuous flow of electrons from
emitter to collector through the base begins.
The emitter current (Ie) is more than the collector current
(Ic). The base current is the difference between leand Ic
and is proportional to the number of electron-hole
combination in the base. The emitter current is given by
Ie = Ic+ Ib
The base in the transistor thin and lightly doped
The base in the transistor is thin and lightly doped. This is
because the number of carriers in the base should be small
so that only small combination of hole-electron may take
place in it and the collector current may not decrease
considerably {i.e. the charge carriers constituting emitter
current may pass on comfortably to the collector region).
6.14.
Amplifier
A device which increases the amplitude of the input signal
is called amplifier.
according to eqn. (1), collector current (Ic) also increases.
Therefore, the voltage drop across RL (= ICRL) increases.
According to eqn. (2), the collector voltage or output
voltage (Vo) decreases. Since collector is connected to the
positive terminal of the battery (VCE) so decrease in V0
means that the collector voltage becomes less positive. In
other words, amplified negative signal is obtained across
the output.
Similarly, during negative half cycle of input signal, the
forward bias of emitter-base junction decreases. As a result
of this, emitter current (Ie) and hence collector current (Ic)
decreases. Therefore, voltage drop across RL(= ICRL)
decreases. Hence according to eqn. (2), the output or
collector voltage (Vo) increases. Since collector is
connected to the positive terminal of the battery (VCE) SO
increase in Vo means that the collector voltage becomes
more positive. Thus, an amplified positive signal is
obtained across the output.
Note:The output voltage decreases with the increase in
input signal and hence there is a phase difference of 1800
between the
input and output signals.
Current gain Resistance gain and Voltage gain of a
transistor
Current gain: It is the ratio of small change in collector
current
to the small change in base-current.
ie.
 Ic 

 Ib 
ac  
Resistance gain: It is defined as the ratio of output
resistance to the input resistance.
R
Re sis tan ce gain  0
Ri
Voltage gain: It is defined as the ratio of small change in
the output voltage to the small change in input voltage.
Av 
V0
Vi
Q. Calculate the emitter current for
β=50 and base current 5μA
The input signal to be amplified is applied across the input
circuit (base-emitter circuit). The input circuit is forward
biased using a battery of e.m.f. = VEB volt.
The amplified output signal is taken across the load
resistance in the output circuit (collector-emitter circuit).
The output circuit is reverse biased using a battery of e.m.f.
= VCE volts.
The emitter-current is Ie= Ic + Ib
------(1)
the output or collector voltage
V0 = VCE- ICRL -------(2)
Ans.
Since β =50 and I b = 5μA
I
Now, β = c  Ic = β×Ib = 50×5 = 250 μA = 0.25 mA
Ib
Also, Ie = Ic + Ib = 250+5 = 255μA = 0.255mA
During positive half cycle of input signal, due to increased
forward bias, emitter current (Ie) increases and hence
104
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Oscillator (Transistor as oscillator)
Oscillator is a device which delivers a.c. output waveform
of desired frequency from d.c power even without input
signal excitation.
L-C circuit producing L-C oscillations consists of an
inductor of inductance L and a capacitor of variable
capacitance C. It can be
connected between the emitter-base circuit. An inductor of
inductance L‘ is connected in the collector - emitter circuit
through a battery and a tapping key (K). Inductors L and L‘
are inductively coupled.
When key K is closed, collector current begins to flow
through the coil L‘ and magnetic flux linked with coil L‘
increases (i.e. changes).
Since coil L is inductively
coupled with L‘, induced e.m.f. is set up across the coil L.
The emitter-base junction is forward biased. The emitter
current Ieincreases which in turn increases the collector
current Ie = Ic + Ib. With the increase in collector current,
magnetic flux linked with coil L‟ also increases. This
increases the e.m.f. induced in the coil L. The increased
induced e.m.f. increases the forward bias of emitter-base
junction. Hence emitter current is further increased which
in turn increases the collector current. The process of
increasing the collector current continues till the magnetic
flux linked with coil L‟ becomes maximum(i.e. constant).
At this stage, the induced e.m.f. in coil L becomes zero.
The emitter current starts decreasing, e. m.f. is again
induced in the coil L but in the opposite direction. It
opposes the emitter current and hence collector current
ultimately decreases to zero. The change in magnetic flux
linked with coil L‟ stops and hence induced e.m.f. in the
coil L becomes zero. At this stage, the capacitor gets
discharged through coil L but now in the opposite direction.
Now the emitter current and hence collector current
increases but now in the opposite direction. This process
repeats and the collector current oscillates between
maximum and minimum values.
frequency is given by  
6.15.Advantages
and
disadvantages
of
semiconductor devices over a vacuum diode
Advantages
1. Semiconductor devices are very small in size and are
more compact.
2. In vacuum tubes, heating of filament is required . In
semiconductor devices no heating is required and the
circuit begins to operate as soon as it is switched on.
3. Semiconductor devices require low voltage for their
operation as compared to the vacuum tube. So lot of
electrical power is saved.
4. Semiconductor devices do not produce any humming
noise which is large in case of vacuum tube.
5.
Semiconductor devices have longer life than the
vacuum tube.
6. semiconductor devices are cheap as compared to the
vacuum tubes.
Disadvantages
1. Semiconductor devices are heat sensitive. They get
damaged due to overheating and high voltages. So they
have to be housed in a controlled temperature room.
2. The noise level in semiconductor devices is very high.
3. Semiconductor devices have poor response in high
frequency range.
ADDITIONAL QUESTIONS
Q1. Why is a semiconductor damaged by a strong
current ?
Ans. When strong current passes through a semiconductor,
large amount of heat is produced in it. This heat energy
breaks almost all the covalent bonds in the semiconductor
which becomes damaged.
Q2.Why
is
the
conductivity
of intrinsic
semiconductor increased even at low temperature
with the addition of impurity to it ?
Ans. Impurity added to the intrinsic semiconductor
provides the current carriers even at low temperature.
Hence its conductivity increases.
Q3. How does the conductance of a semiconducting
material change with rise in temperature ?
Ans: It increases.
Q4.
How does the energy gap in an intrinsic
semiconductor vary, when doped with a pentavalent
impurity ?
Ans. The energy gap will be reduced.
Q5. What is meant by the term doping of an intrinsic
semiconductor ? How does it affect the conductivity of a
semiconductor ?
1
2 LC
Q6. How resistance varies in semiconductors with
temperature ?
Ans. Resistance of semiconductor varies as inversely
proportional to the temperature of the semiconductor.
105
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Ans. Majority carriers are holes and minority carriers are
electrons.
Q7. What type of charge carriers are there in a p-type
semiconductor ?
Q9. What is a pn junction? What is barrier potential?
Ans. When a semiconductor crystal (germanium or silicon)
is so prepared that one half is p-type and the other n-type,
the contact surface dividing the two halves is called a pn
junction.
The diffusion of electrons and holes establishes a potential
difference across the junction. This is called potential
barrier or junction barrier V0
Q8. What type of charge carriers are there in a n-type
semiconductor ?
Ans. Majority carriers are electrons and minority carriers
are holes.
Q10. What are forward biasing and reverse biasing of a
junction diode?
How is the barrier potential affected during forward
biasing?
Ans. Forward biasing: When external voltage applied to
the junction is in such a direction that it cancels the
potential barrier, thus permitting current flow, it is called
forward biasing.
Barrier
potential
decreases.
LOGIC GATES
POWER QUESTIONS
Q1. Explain the OR gate, AND gate and NOT gate.
Ans. (i) The OR gate It has has two or more inputs but only one output. It is called OR gate because the output is high if any or
all the inputs are high.
The relation between the output (Y) and the inputs (A and B) is given by the Boolean expression.
Y = A + B and is read as ―Y equals A OR B‖.
(ii) The AND gate. AND gate has two inputs and only one output.
The relation between inputs (A and B) and the output is given by Y = A.B and is read as ‗Y equals A AND B‘.
(iii) The NOT gate.
The NOT has only one input and The relation between input (A) and
only one output. Output (Y) is given by Boolean expression
Y  A
The NOT gate is also known as inverter gate.
Q2. What is a digital circuit?
Ans. An electronic circuit that handles only a digital signal is called a digital circuit.
Q3. What is a logic gate? What are the different types of logic gates?
Ans. A digital circuit with one or more input signals but only one output is called a logic gate
106
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Q 4. Explain the formation of NAND gate and NOR gate. Draw their symbolic representations and the corresponding
truth tables.
Ans. The NAND gate. The NAND gate is basically a NOT-AND gate. The logic gate in which the output of the AND gate is
given to the input of NOT gate is called the NAND gate.
The Boolean expression for the NAND gate is given by
Y  A. B
The NOR gate. The NOR gate is a NOT-OR logic gate.
The logic gate in which the output of the OR gate is given to the input of NOT gate is called the NOR gate.
Boolean exp ression for NAND gate
Y  A B
Q5. What do you mean by encoding and decoding ?
Ans. Digital circuits understand only binary numbers i.e. input to a digital circuit must be in binary form. Encoding means
changing decimal numbers to a binary form. In decoding ,the digital information is converted back into decimal numbers.
Q536. Give two advantages of digital circuits over the analog circuits.
Ans. The two advantages of digital circuits over analog circuits are :
(/) Information can be stored for short and indefinite period.
(ii) Digital circuits are very accurate.
Q6 .
What are the limitations of digital circuits ?
Ans. The following are the limitations of digital circuits as compared to analog circuits : (i) Most of ―real world‖ information
dealing with time, speed, pressure and position
measurements are analog in nature. (ii) Analog processing is generally simpler and faster.
Q7. What is a truth table ?
Ans. The action a logic gate is usually summarized in the form of a truth table. It defines the output of a logic gate for all possible
combinations of inputs.
Q8. What is an inverter?
Ans. The NOT gate is called inverter. It has only one input and one output. It is called inverter as the input and output are always
opposite.
107
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Analog signal
An electrical signal which varies continuously over time is
known as analog signal.
UNIT X
Communication Systems
Communication
The process of sending and receiving information is
known as communication.
FAX
Digital signal
Digital signals are discontinuous and discrete signals
having only binary variations 1 and 0 with time.
Modulator
A device used to modulate a high frequency carrier signal
is known as modulator.
Modem and its function
Modem stands for modulator and demodulator. It converts
digital signals to analog signals and vice versa.
6.16.
The electronic reproduction of a document at a distant
place is known as FAX.
Functions performed by modem
A modem converts analog signal into digital or digital
signal into analog i.e. it performs the functions of
modulation and demodulation.

Type of modulation
required for TV
broadcast Frequency modulation (FM).
 Frequency of carrier wave
It should be very high i.e. in kHz to GHz.Type of
modulation scheme preferred for digital communication.
Pulse modulation.
Types of communication systems according to mode of
transmission
Cable communication, optical fiber communication.
Modulation
The process of variation of some characteristic of a high
frequency carrier wave in accordance with the
instantaneous value of audio wave is known as modulation.
Need of modulation
For differentiation of a signal, a particular portion of the
frequency spectrum has to be assigned to that
signal so that only the desired one is selected and the rest
are rejected. Similarly, a carrier wave of high frequency
can not be used to convey information without modulation
because the carrier wave has fixed parameter i.e., its own
amplitude or frequency or phase remain constant until or
unless any of these parameters is selected to vary in
accordance with information signal. Thus, modulation is
necessary for a low frequency signal when it is to be sent to
a distant place so that the information may not die out in
the way.
Different types of modulation and their advantages and
disadvantages
(i) Amplitude Modulation(AM):The process of changing
the amplitude of a carrier wave in accordance with the
amplitude of the audio frequency signal (AF) is known as
amplitude modulation (AM.)
When a low frequency modulating signal (figure (a)) and a
high frequency carrier wave (figure (b)) are superimposed
to produce amplitude modulation, the resultant modulated
wave is available as shown in figure (c).
Advantages of Amplitude Modulation;
(1)
Amplitude Modulation is an easier method for
transmission
and receiving of speech signals.
(2) It requires simple and inexpensive receivers.
(3) It is considered to be a fairly efficient system of
modulation.
Demodulation
The process of extracting the audio signal from the
modulated wave is known as demodulation.
108
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
Disadvantages/Limitations/Drawbacks of Amplitude
modulation
(1) Amplitude modulation is more likely to suffer from
noise.
(2) The equipment is more complex.
(3)
Cost of such transmitters and receivers becomes
practically more.
(ii) Frequency Modulation(F.M):The process of changing
the frequency of a carrier wave in accordance with the
audio frequency signal is known as frequency modulation
(FM).
The process of frequency modulation changes the
frequency of the carrier wave without changing the
amplitude of the carrier wave.
Ans. PAM, PWM and PPM are analog modulations
whereas PCM is digital modulation.
PAM, PWM and PPM are analog
Because in all these modulations, the modulated signal has
some characteristic like amplitude, width or position, which
is variable and proportional to the instantaneous
modulating voltage.
Bandwidths of an A.M. radio station and an F.M. radio
station
The bandwidth of the amplitude modulation radio station is
generally 10 kHz. The bandwidth of the FM radio station is
generally 150 kHz.
Advantages of FM
1. FM is inherently and practically free from the effects of
noise.
2. FM receivers can further be improved with the help of
limiters which controls the noise level.
4. All the transmitted power is useful in FM.
5. In FM, it is possible to operate many independent
transmitters on same frequency without interference.
ADDITIONAL QUESTIONS
Limitation/disadvantages/drawbacks of FM
1. About 10 times wider channel is required by FM as
compared to AM.
2. Area of reception for FM is much smaller than for AM.
3. FM receivers and transmitters are very complex and
costly.
Q. What are digital and analog communications?
Ans. The type of communication in which digital signals
are used is known as digital communication. If the circuit
uses analog signal, it is known as analog communication.
Q. What is the band width required for FM radio
stations ?
Ans. 150 KHz
6.17.
Pulse modulation and its types
It is a type of modulation in which some parameter of the
pulse train carrier is varied in accordance with the
instantaneous value of the modulating signal.
(i) Pulse Amplitude Modulation (PAM). Here, the
amplitude of the pulses of the carrier pulse train is varied in
accordance with the instantaneous value of the modulating
signal.
Q. How is speech reproduced in a receiver ?
Ans. The extracted audio frequency signal from modulator
is fed to the audio frequency amplifier where it is
amplified. The amplified audio frequency signal is given to
the speaker, which converts it into original speech (i.e.,
sound wave).
Q. Explain with the help of block diagram, how the
process of modulation is carried out in radio broadcast.
Ans.The transmitter consists of :
(i) a microphone to convert voice signal into electrical
signal
(ii) a modulator to mount the low frequency signal on the
high frequency carrier wave
(iii) amplifier to make the signal stronger and (iv) an
antena to transmit the modulated wave (converted by the
modulator)
The receiver. In the radio communication or wireless
communication, the receiver consists of:
(i) a pick up antenna to pick the signal,
(ii) a demodulator, to separate the low frequency
audio signal from the modulated signal, (III) an amplifier, to
boost up suitably the audio signal, and (iv) the transducer,
like loud speaker to convert the audio signal (in the form of
electrical pulses) into sound waves.
(ii) Pulse Width/Duration Modulation(PWM/PDM).
Here, the width or duration of the pulses of the carrier pulse
train is varied in accordance with the instantaneous value of
the modulating signal.
Q. What are the advantages of digital communication?
Ans. (i) Reproduction of original signals is more accurate.
(ii) Less distortions.
(iii) Pulse Position Modulation (PPM).
Here, the position of the pulses of the carrier pulse train is
varied in accordance with the instantaneous value of the
modulating signal.
The radio waves which are reflected back to the earth by
ionosphere are known as sky waves.
The mode of propagation of sky waves is known as sky
wave propagation. In this type of propagation, radio
waves transmitted by a transmitting antenna are directed
towards the ionosphere. The radio waves having frequency
range
Q. Which of the following are analog and digital
modulations PAM, PWM , PPM and PCM ?
6.18.
Sky waves and its propagation
109
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
2 MHz to 10 MHz are reflected back by the ionosphere.
The successive reflections of these radio waves at the
earth‘s surface and the ionosphere make it possible to
transmit these waves from one part to any other part of the
globe.
Radio waves having frequency nearly greater than 10 MHz
penetrate the ionosphere and are not reflected back by the
ionosphere. Therefore, these waves are not propagated
through this mode of propagation. Sky wave propagation is
useful for very long distance radio communication
Intercept the signals from the transmitting antenna. For
example, for good and large coverage of TV programme,
tall antennas must be used for transmitting the TV signals.
Micro wave transmission on the surface of earth is possible
if the transmitting and the receiving antennas are in a lineof-sight.
Satellite communication
Critical frequency
Critical frequency (CF) is defined as the highest frequency
that is returned to the earth by the considered layer of the
ionosphere after having been sent straight (normally) to it.
Fading
Fading is defined as the variation in the strength of a
signal at a receiver due to interference of waves.
SID and Skip Zone
Sudden ionospheric disturbances (SID) is defined as the
abrupt change in the properties o; ionosphere due to solar
flares.
Skip zone :The region over which no signal is received
either from ground wave propagation or the sky wave
propagation is known as skip zone.
7.20.Space wave communication
Ans. The propagation of very high frequency (VHF), ultra
high frequency (UHF) and microwaves is not possible
through ground waves and sky waves. This is because,
VHF and UHF waves are almost absorbed by the surface of
the earth after travelling a small distance.
These high frequency waves (above 30 MHz) called space
waves can be transmitted from transmitting to receiving
antenna through a mode known as space wave propagation.
This mode of propagation is limited to the line of sight.
There must be no obstruction (like curvature of the earth or
mountains or huge buildings) in between the transmitting
and receiving antennas.
The transmission of these waves is possible due to their
refraction in troposphere. Since the temperature
troposphere decreases with the height, so the refractive
index of this layer of the atmosphere decreases with height.
The space waves are transmitted by the transmitting
antenna towards the receiving antenna. These waves suffer
refraction in the troposphere and bend slowly towards the
earth‘s surface. Due to this refraction,
the line of sight distance is increased. This modified line of
sight
distance can be calculated for the given height of the
transmitting antenna.
Very high frequency signals can be transmitted without
losses in air only if the receiving antenna directly
Satellite communication is like the line-of-sight microwave
transmission. In this case, a beam of modulated micro
waves is projected towards the satellite.
The signals are received by a device known as
transponder fitted on the satellite. This device re-transmit
these signals after amplification towards the earth. These
signals are received by the receiving stations on the earth.
The signals received are very weak, so they are amplified at
the
receiving
stations
and
then
they
are
broadcasted/telecasted. Thus, a communication satellite
acts as a big microwave repeater in the sky.
The requirement for the line-of-sight transmission of
microwaves is that the transmitting and receiving antennas
must always be in the sight of each other. This can be
possible only if the communication satellite is always at a
fixed point with respect to the earth. In other words, the
satellite used as a repeater must be at rest with respect to
the earth.
A geo-synchronous satellite can establish a communication
link over a large part of the earth but a single satellite
cannot cover the whole part of the earth as the curvature of
the earth keeps a large part of the earth out of sight. At least
three satellites put in the synchronous orbit are required to
provide the communication link over the whole part of the
earth.
Disadvantages (i) Since the information transmitted
through satellite can be heard/caught by every body, so
satellite communication is not good in respect of security
and privacy point of view.(ii) The cost of placing geosynchronous satellites in the orbit is very high.
ADDITIONAL QUESTIONS
Q. Long distance radio broad-cast use short wave
bands. Why ?
Ans. Long distance radio broadcast is possible due the
reflection of the radio waves by the ionosphere. As
110
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
ionosphere reflects only the waves having a frequency
range 2 MHz to 30 MHz (short wave band), so long
distance radio broadcast use short wave bands.
Q. Sky wave propagation is not possible for very high
frequency radio waves. Explain why ?
Ans. Sky wave propagation is due to the reflection of radio
waves by the ionosphere. Ionosphere is transparent to very
high frequency waves and hence these waves cannot be
reflected by the ionosphere.
Q.Geo-stationary satellite is used for the line-of-sight
transmission of microwaves. Comment.
Ans.The requirement for the Line-of-sight transmission of
microwaves is that the transmitting and the receiving
antennas must always be in the sight of each other. Since
the geo-stationary satellite is at rest with respect to the
earth, so the transmitted signal is projected towards the
geo-stationary satellite. The geo-stationary satellite reflects
this signal towards the receiving antenna.
Q. What are the bandwidths of an A.M. radio station.
Ans.The bandwidth of the amplitude modulation radio
station is generally 10 kHz.
Q. Mention the ranges of carrier frequencies used for
an A.M. radio station.
Ans. Amplitude modulation radio station uses carrier
frequencies between 530 kHz to 1700 kHz, Each radio
station‘s carrier frequency must be separated from the next
stations carrier frequency on either side by 10 kHz to avoid
the overlapping of the signals transmitted by these stations.
Q. Ground wave propagation mode is not suit-able for
the propagation of very high fre-quency e.m. waves to
long distances. Ex-plain why ?
Ans. The energy of the electromagnetic waves decreases as
they travel over the surface of the earth due to the
conductivity and permittivity of the surface of the earth.
The decrease in the energy of the waves increases with the
increase in the frequency of the electromagnetic wave.
Therefore, the energy of the very high frequency e.m.
waves decreases very fast over a small distance and hence
they can not travel long distances over the earth‘s surface.
Q. What should be the frequency of carrier wave?
Ans. It should be very high i.e. in kHz to GHz.
Q. What is holography ?
Ans.The technique for recording and reproducing three
dimensional (3D) image of an object with use of lasers is
known as holography. This technique is used in
cinematography, computer images and holography
microscopes.
Q.
What
is
a
transponder?
Ans. A device fitted on the satellite which receives the
signal and retransmit it after amplification.
Q. What is the requirement of transmitting
microwaves from one position to another on the earth ?
Ans. The transmitting and receiving antennas must be in
line of sight.
Q. What is a geo-stationary satellite ?
Ans.A satellite which is at rest with respect to the earth.
This satellite remains fixed at a point w.r.t. the earth.
Q. What should be the height of geo-synchronous orbit
from the surface of the earth ?
Ans. About 36,000 km from the equator.
Q How many geo synchronous satellites are required to
provide the communication link over the whole part of
the earth ?
Ans. Minimum three.
Q. What is remote sensing ?
Ans. The technique to collect information about an object
without actually touching it.
Q. Name the type of radio waves propagation involved
when TV signals broadcast by a tall antenna are
intercepted directly by the receiver antenna.
Ans.
Space
wave
propagation.
Q.Why is the transmission of signal using ground waves
restricted to frequency about 1500 kHz ?
OR
Why ground wave propagation is not suitable for high
frequency ?
Ans. This is because signals having frequency more than
1500 kHz are greatly absorbed by the surface of earth over
a small distance.
111
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
112
BRILLIANT SUCCESS
Higher Secondary Exam -2011 CRACKER (PHYSICS)
113