TARGET IIT-JEE-2015
PART TEST-04
XII-PASS
ANSWER WITH SOLUTION
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PHYSICS
1.
A

mA
I
O
I
O
1
 A  B  A B 
mB O A IB
IB O A 3
2.
B
mA 
But
f

mB 
f  uA and
f  uB
1
7.5  uB

3   7.5   30 
For minimum deviation i = e. Therefore m  2i  A or A  m  2 i or
3.
 A  m 
sin 

 2   sini


Now
A
A
sin
sin
2
2
(C)
2 1 2  1


u
R
/
 A  m   i
2
A
1
 60 
Sin i =  sin 2  2 sin  2   2 sin30 
2


where 1  1,  2  1.6,   50 cm & R = + 12 cm.
/  
160
cm
3
the v irtual object for the second surface so that    2R  is the object distance for this surface.
1 2 1  2



u
R
4.
  9.5 cm
D
The amplitudes of the two coherent waves will be A 1 = 2A and A2 = A.
therefore
2
2
Imax  A 1  A 2 
 2A  A 

 
  9 Hence the correct choice is D
Imin  A1  A 2 
 2A  A 
5.
C
White light consists of colours between violet source S 1 will decrease. The intensities of the two
virtual sources will be different. Hence the correct is C
6.
(B) path diff = (n1L1 – n2L2) : Phase diff = (2 / ) (n1L1 – n2L2).
C
We know that in case of a convexlens when object is placed at C / , the image is obtained at C.
This situation is represented in the graph by the point corresponding to u = – 10 cm, v = 10 cm.
7.
Therefore R  10 
R
 5 cm  f Lens formula is
2

(for maximum error in f)
u
L
(A)
x
9.
df
0.1
0.1


 df  25  0.1 2  0.01  0.05
2
2
25 10 
10 
v
L
8.
1 1 1
df
d du
   2  2  2
f v u
f
v
u
I
v
u
v
Here, (v / u) = m1 and (u / v) = m2
 m1 – m2 = (v / u) – (u / v)
= {(v2– u2) / uv} = [{(v – u) (v + u)} / uv]
= [x × (1 / f )] [ f = {uv / (u + v)} and v – u = x]
 f = {x / (m1 – m2)}
(B) The power of the lens is given by
P1 = (1 / f1) = (µ – 1) [(1 / R) + (1 / R)] = (1.5 – 1) (2 / R) = (1 / R)
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the power of each water lens (plano-concave)
P2 = P3 = (1 / f2) = (µ – 1) (1 / R) = {(4 / 3) – 1}(–1 / R) = – (1 / 3R)
The power of combination
P = P1 + P2 + P3 = (1 / R) – [(1 / 3R) + (1 / 3R)] = (1 / 3R) = {1 / (3 × 10)}
= (1 / 30 cm) = (1 / 0.3) dioptre = 3.33 dioptre
10. (A) Here (90 – r) should be just greater than critical angle. Thus
 cos r = (1 / 1µ2) = (µ1 / µ2)
1µ2 = [1 / (sin (90 – r)] = (1 / cos r)
2
 (1 – sin r) = (µ1 / µ2)
.....(1)
From figure, (sin / sin r) = (µ2 / µ1)  sin r = (µ1 / µ2) sin  .......(2) F r o m
eqs. (1) and (2), we get
[1 – {(µ1 / µ2) sin  ] = (µ1 / µ2)
 [1 – (µ1 / µ2)2 sin2  ] = (µ1 / µ2)2
Solving we get sin  = [(µ2 / µ1)2 –1]
11. Ans. B
Let d = the diameter of the tube.
 / 4  24.1  0.3 d , and 3 / 4  74.1 0.3d
3 / 4  4.1  0.3 d or


 50 or
 100 cm 
2
2
0.3d    / 4  24.1 cm   25  24.1 cm  0.9 cm
or d = 3 cm
12. Ans. D
V
2V
1


 
L


L
2L
Frequency =  
where  = the velocity of transverse waves on the string.
2
13. Ans. A
In a wave equaiton, x and t must be related in the form (x–t).
We rewrite the given equation as
y
1
1

1   x  t  .

For t = 0, this becomes y = 1  x 2 , as given
2
1
2
for t = 2 this becoemes y = 1   x  2 



1
1   x  12  


2   1 or   0.5 m / s
y = a coskx cos t  sinkx sin t  is given
14. Ans. C
As y =0 ato x = 0 for all values of t(node), the cos kx term must vanish in the sum of the two waves
15. Ans. D
Initially, 260 
1
2
T1
, T1  50.7 g  507 N .
m
= (0.0075 m³) (10³ kg/m³) (10 m/s²)
New tension = T 2 = (507 – 75) N = 432 N
or
16. Ans. C
 I1 
40 10 = log10  I 
0
When the mass is submerged, upthrust
= 75 N
n=
1 T2
2 m
n
o r 260 
T2

T1
432 144 12


507 169 13
n = 240 Hz.

I1
 10 4
I0
........(i)
 I2 
Also, 20  10log10  I 
0
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
I2
 102
I0
.............(ii)

I2
r2
 10 2  12
I1
r2
r2  100r12  r2  10m r1  1m 
17. Ans. A
There will be no effect of magnetic force on time period because the magnetic force will be
perpendicular to the inclined plane.
q, m

18. C
2
T  2
2
m
m
2 m
4 m
 2
T2 
or k 

k eff.
2k or
2k
T2
2
 22 
2
 12
2
7 
 22 

k
 6
 59.15 Nm 1

4
 7 
19. B
As here two masses are connected by two springs, this problem is equivalent to the oscillation of
a reduced mass  of a spring effective spring constant.
T  2

k eff. Here,
1 k
1 2k

2
 m  2
20. D

m1m2
m

m1  m2
2 k eff  k1  k 2  2k 
n
2 k eff.
1 2k

2
2 
2 m
1 k
1 0.1 1

 Hz
 m  0.1 
Here, m = 4kg, k = 800 Nm1 ; E = 4 J
Is SHM, total energy is E =

4

1
 800  A 2
2
1 2
kA , where A is the amplitude of oscillation.
2
2
Maximum acceleration, amax.   A 
k
A
m

  

k 

m 
800 Nm 1
 0.1 m  20 ms 2
4 kg
21. B
Let m be the mass of each particle, then 1   h / m1  and  2   gh / m2  , where 1 and 2


 

m1  m 2 1  2
CM 

2m
2
 

 
  2
1c  1   CM  1
2

Velcoity of A w.r.t. C frame is,

| 1c |
 
1   2
h
h
2
 

m | 1c | m h 1
1 
 2
2
m 1  2
2

|  2c |
So, required wavelength is
21 2
12   22
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22.
(c)
4 2 l
T2
T  2 l/g  T 2  4  2 l/g  g 
Here % error in l =
1mm
0 .1
 100 
 100  0 .1 %
100 cm
100
0.1
and % error in T = 2  100  100
 0 .05 %
 % error in g = % error in l + 2(% error in T)
 0 .1  2  0 .05 = 0.2 %
23. (b)
24.
(b)
V 
4 3
r
3
 % error in volume
 3  % error in radius.

25.
3  0.1
 100
5.3
(c)
Volume of cylinder V  r 2 l
Percentage error in volume
V
2 r
l
 100 
 100   100
V
r
l
0.01
0. 1


 2 
 100 
 100   (1  2)%
2
.
0
5
.
0


= 3%
26. A
N2
3
 7x

e  
N1 100
We have N1  N0 e7 & N2  N0 ex 

  x  7   ln
100
100
 2.3log10
3
3
0.693 0.693

But   T
12.5
1/2

100
 x 7 

; e
3
= 2.3 [ 2 – 0.4771]= 2.3 × 1.5229
x–7 =
12.5
12.5
 2.3  1.5229 =
 1.52  63
0.693
3
 x= (63 + 7) years = 70 years
27. C
We have
K.E. = k T = 1.38 × 10 –23 × 300 J
=
1.38  1023  300
1.38  3
eV 
 10 2 eV
19
1.6
1.6  10
= 0.026 eV
28. C
Binding energy
= [ZMP   A  Z  MN  M]c 2
2
=  8MP  17  8  MN  M c
= 8MP  9MN  M c 2
= 8MP  9MN  M0  c 2
3
RhC  V e.v. 
4
RhC 
29. C
Enery released = (B.E. of product – BE of reactant)
(80 × 7 + 120 ×8 – 200 × 6.5) = 220 MeV
30. C
1
 1
First excitation energy = RhC  2  2
2
1
3

  RhC 4 

4V
e.v.
3
CHEMISTRY
31.
A
f1 z12 n32 n1  3 n2  3 22 23 32

, 

Solution:  3  2
f2 n1 z 2 z1  2 z 2  1 33 1 27
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32.
C
Solution: Maximum number of electron having same spin quantum number is equal to the number of orbitals  2  1
33.
C
Solution: Magnetic moment =

n(n  2)  3.873
 25Mn  [Ar]3d5 4s2
therfore Mn should be in + 4
4s
3d2
34.
35.
36.
37.
38.
number of unpaired electron n  3
A
A
B
D
B
Solution:
2NO(g)
+
moles 2
2 – 2 × 0.5
1
nf = 1 + 1
O2 (g)
0.5
0
0

2NO2 (g)
0
2 × 0.5
1
ni = 2 + 0.5
 n  (2.5  2)  0.5 moles
 changein pressure
p 
nRT
1 300
 0.5 

 2 atm.
V
12 6.25
39.
C
Solution: c
S1: Na2O2  MgO  ZnO  P4O10 : as non-metallic character increases the acidic character increases. MgO,
Na2O are basic, ZnO amphoteric and P4O10 acidic.
S2: Na  Si  Al  Mg : Mg has higher than Na due to small size and higher nuclear charge. Mg has higher than Al
because of ns2 configuration (has extra stability and high electron penetration power of s-subshell electrons) and
Si has higher than Al because of higher nuclear charge and small size. IE1 : Na = 496, Al = 577, Mg = 737 and
Si = 786 kJ/mole
S3: There is more interelectronic repulsion in 2p-subshell of fluorine than chlorine (3p). So extra electron will be
added easily in 3p-subshell of chlorine as compared to 2p-subshell of fluorine. Down the group electron affinity
values generally decreases with increasing atomic number due to increase in atomic size. So Cl > F > Br
S4: Isoelectronic series of ion; all have the xenon electron configuration .
Ionic radius 
1
nuclear charge
Atomic number : Te = 52; I = 53; Cs = 55; Ba = 56.
40.
A
Solution: As non-metallic character of element attached to oxygen atom increases, the difference between the
electronegativity values of element and oxygen decreases and the acid character of oxides increases and
vice-versa.
41.
C
Solution:
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42.
A
Solution: C  O  1.43Å; C  O 1.23 Å ; C  O 1.09Å

CO  C  O; partial triple bond character; bond length is inter mediate of C = O and C  O bonds CO2  O  C  O;
double bond length
43.
D
Solution: (d)
(a)
Bond order 
1
bond length
Bond order O 2  2.5, O 2  2 O2  1.5
So correct of bond length is O 2  O 2  O 2 .
(c)
22
 0. So He2 does not exist.
2
In all these molecules all electrons are paired in molecular orbitals. So they are all diamagnetic
(d)
F2 MOT configuration: 1s2 *1s2 2s2  * 2s2 2pz2 2pz2  * 2py2  * 2pz2  * 2pz2
(b)
Bond order for He2 
44.
D
Solution: (d)
Adjacent bonds, thereby distoring the structure as doubly bonded oxygen atom require more space than a
single bonding pair and repel.
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45.
C
Solution: c
46.
B
Solution: Cyclohexane (C6H12) and carbontetrachloride (CCl4) both are non-polar compounds so there exist London
dispersion force between the constituent molecules.
47.
B
Solution: b
(2) KO2  2H2O  KOH  H2O2  1/ 2 O 2
(3) 4KO2  2CO2  2K 2CO3  3O2
A
Solution: S1: In concentrated solution unpaired electrons pair up together and become dimagnetic.
S2 and S3 are correct statements
48.
S4: Be(OH)2  2OH  [Be(OH)4 ]2 beryllate ion
Be(OH)2  2HCl  2H2O  [Be(OH2 )4 ]Cl2
49.
C
Solution: S1: True statement - salt of weak base and strong acid.
S2: As we move down the group from Li to Cs the strength of metallic bonding decreases as size of atom
increases. Due to this three occurs a decrease in close packing of atoms in crystal lattice from Li to Cs and thus
the softness increases from top to bottom.
S3: It is hemihydrate of CaSO4(2CaSO4.H2O)
S4: KHCO3 is soluble in water and cannot be precipitated like NaHCO3
50.
D
Solution: d
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(a) P4  3NaOH  3H2O  PH3  3NaH2PO 2

(b) 4S  6NaOH 
 Na 2S2O3  2Na 2S  3 H2O
(c) 3 Cl2  6 NaOH  5 NaCl  NaClO3  3 H2O
(d) 2B  6 NaOH  2 Na3BO3  3 H2
51.
C
Solution: c
(a)
It is a weak monobasic acid soluble in water and in aqueous solution the boron atom completes its octet by
(b)
(c)
52.
(d)
A
accepting OH– from water molecules: B(OH)3  2H2O  [B(OH)4 ]  H3O 
Only TI3+ acts as oxidising agent on account of inert pair effect
In the solid state, the B(OH)3 units are hydrogen bonded together into two dimensional sheets with almost
hexagonal symmetry
B(OEt)3 imparts green colour to burner flame.
Solution: NH4ClO 4  HNO3  HClO 4  NH4NO3

2NH4NO3 
 2N2O  4H2O
53.
D
Solution: d
(1)
(2)
(3)
(4)
54.
A
Solution: a
55.
NH3 has higher boiling point than AsH3 and PH3 on account of H-bonding. SbH3 has highest boiling point on
account of highest molecular weight.
As the size of element increases, the strength of M–H bond decreases and thus stability decreases.
Basicity decreases down the group. (Drago rule).
HEH angle (°) : NH3 = 107.8 ; PH3 = 93.6, AsH3 = 91.8 ; SbH3 = 91.3
Solution: d
(a)
(b)
(c)
(d)
Disproportionation is slow at the ordinary temperature for OCl , it is fast for OBr  and very fast for OI–
E° for the 1/2 F2 | F– electrode = + 2.9 V
HI stronger acid than HBr because of its low bond dissociation enthalpy
Because of low bond dissociation energy and high hydration energy of F2, F2 acts as strong oxidising agent
and oxidising power decrease down the group. E° 1/2 F2 | F = + 2.9 V
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-9Cuttack Study Center Ph.:-0671-6544599,2314999,8456887020
Amogh Classes–PART TEST-04-TARGET IIT-JEE-2015
56.
C
Solution:
57.
A
Solution: Sulphite gives SO 2 with H2SO 4 which turns acidified K 2Cr2O7 green according to following reaction
K 2Cr2O7  H2SO4  3SO2  K 2SO 4  Cr2 (SO 4 )3 (green)  H2O.H2S also gives green coloured solution with acidi-
58.
fied K 2Cr2O7 solution but a slightly coloured precipitate of sulphur is also obtained. The H2S gas has rotten egg
smell.
D
2
HCl
NaOH
 MnCl2 
 MnO2  (Brown / black)
Solution: Mn  H2S  MnS  (buff / light pink coloured) 
[O]
59.
fusion
MnO2  2KNO3  Na2CO3 
 Na 2MnO 4 (green compound)  KNO 2  2NaOH  CO2
A
[O]
Solution: MnO 4 / H  [O] : Fe2 
 Fe3  [Fe(CN)6 ]4   Fe 4 [Fe(CN)6 ]3
60.
D
Solution: d
(b)
Cu2  4NH4 OH  [Cu(NH3 )4 ]2 (deep blue)  4H2O
Bi3  3NH4OH  Bi(OH)3  (white)  3NH4
(c)
Cu2  [Fe(CN)6 ]4   Cu2 [Fe(CN)6 ]  (chocolate brown)
Bi3  [Fe(CN)6 ]4  no change observed
MATHEMATICS
61.
66.
71.
B
D
A
62.
67.
72.
B
A
C
63.
68.
73.
C
A
C
64.
69.
74.
C
A
B
65.
70.
75.
B
A
C
76.
B
77.
D
78.
C
79.
C
80.
C
81.
86.
B
A
82.
87.
C
A
83.
88.
C
C
84.
89.
C
B
85.
90.
B
D
AMOGH CLASSES:- Head Office: Plot No.-1189, Nilakantha Nagar, Besides New Passport Office, Devray College Road, Nayapalli, BBSR-12,
Odisha Ph:-2565949,6530219. Damana Study Center. Ph:-0674-2745599,6444399.
- 10 Cuttack Study Center Ph.:-0671-6544599,2314999,8456887020