Document 284751

CSSE
CCE SAMPLE QUESTION
PAPER
FIRST TERM (SA-I)
SCIENCE
(Theory)
(With Solutions)'
CLASS IX
Time Allowed:
3 Hours 1
General Instructions:
.
(i) The question paper comprises of two sections A and
R
You are to attempt
both t1~ sections.
(iI) All questions are compulsonj.
(iii)
There is no overall choice. However, internal choice has been provided in all the five q~ons
marks category. Only one option in each questions is to be attemiJied.
(iv)
(v)
(vi)
(viI)
(viiI)
(ix)
f'
All questions of Sectio~ A and all questions of Section B are to be attempted
QU!!Stion numbers
or one"sentence. ,~
Q~tion
I to 3 in Section
".
v·
" .
to
19 are
'.
separately.
A ar~.an!!,A'narkquestions. These are to be ansr1J;f!l1.,"!. ane-word
...:. ~
'
....,
.. - .' "" -
numbers 4 'loJ :Ue two marksquestioris,
Question numbers.8
of five
to be.answered
in, qbput '30 wo~ds ~d,.
three marks questions; to be answered
m-abaut 50 Words each. ,
Question numbers 2010/4 are five"!f1rks questiOns, to .b~ answered ii~p?ut
70 words, ea~h..
Question numbers 25 to 42 in Sfction B kre multiple choice qw;stions baSedpl1 ptacticol skills. Thcn
question is a aile mark question.
You are to c1Wose one mOst izppropriate response oUt of the Jour
provided to you.
~
SECTION A
Q.l. A gas jar containing air is inverted over another containing NOz gas which is brown
in colour and heavier than air. After some time brown colour is seen in the inverted gas jar
(1)
too. Identify the phenomen~n associated with this observation.
Ans. Diffusion of gases.
Q.2. A passenger in a moving train tosses a coin which falls behind him. State the type.
(1)
of motion of the train.
Ans. The motion of the train is an accelerated motion.
Q.3. Name the plastid involved in conversion of a green tomato to red.
(1)
Ans. Chromoplast.
Q.4. List four reasons to support that water is a compound and not a mixture.
(2)
Ans. The following reasons support that water is a compound and not a mixture :
(i) The composition of water (H20) is fixed. It is in the ratio of 2 : 16 or 1 : 8 for hydrogen
and oxygen.
(ii) Water has fixed melting point (273 K) and boiling point (373 K).
(iii) We cannot separate water into its constituents (H and 0) by physical methods.
(iv) The properties of water are completely different from those of hydrogen and oxygen.
(.A-41 )
. Science-IX
A-42
.l. '.
eSSE
Q.5. A man weighs 600 N on the surface of earth. What would
on the surface of moon ? (Take gearth = 10 m/s2).
=
be his mass and weight
(2)
Ans. As per question weight of a man on the surface of earth Wearth = 600 Nand
10 m/s2
Mass of man on the surface of moon = Mass of man on the surface of earth
gearth
m = Wearth = 600 N = 60 kg
gearth
10 m/s2
Weight of man on the surface of moon Wmoon = m.gmoon·= 1Il.gearth
6 = Wearth
6
between
composition
throughout
mass.
(ii) There
is mixture
no boundary
of its
separation
homogeneous
mixture.
has uniform
(i) Homogeneous
= 600N = 100 N.
6
Q.9. Distinguish
between homogeneous
and heterogeneous
mixtures. Classify the
following mixtures as homogeneous and heterogeneous :
(i) Tincture of iodine
(ii) Smoke
(iii) Brass
(ip) Sugar solution.
(3)
Ans. Distinction between homogeneous and heterogeneous mixtures is given below :
mixture.
between
the constituents of a heterogeneous
a constituentsof
the
uniform
(i) Heterogeneous
(ii)
Therecomposition
is amixture
visible
mixture
throughtout
boundary
does ofnot
its
separation
mass.
have a
Heterogelleolls
Homogelleolls
Q.6. State one feature that is similar and one feature that is dissimilar with respect to .
mitochondria
and plastids.
(2)
Ans. Similar feature: Both have their own DNA and ribosomei to make their own proteins.
Dissimilar feature: Mitochondria are the site of cellular respiration (power house of cell) while
chloroplasts are centre for photosynthesis (food factory of cell).
Q.7. Show'the location of meristematic tissues in a plant diagrammatically.
is responsible
a tree?
Ans.
for the transformation
A-4~
eeE Sample Question Paper
Classification
mixture
of mixtures:
(i) Tmcture of iodine
Which meristem
of the stem of a plant into the trunk when it grows into
(2)
Apical meristem
Intercalary meristem
Homogeneous
Heterogeneous
Brass
Homogeneous
Sugar solution
Homogeneous.
Derive graphically the equation for position-time
in time 't' under uniform acceleration.
(ii) Smoke
(iii)
(iv)
Q.10.
distance's'
Ans. The velocity-time graph has been shown in
At t = D, the initial velocity is u (at point A)
increases to v (at point B) in time t.
Draw perpendicular
lines BC and BE from point
time and velocity axes respectively, so that OA = u,
relation for an object travelling
figure.
and then
B on the
OE = BC
~v.
1,
S____________________
.s
>
v
a
(3)
B
!I
:I
II
.?:
.~~
I
>
t --------------fD
Lateral meristem
Also draw a line AD parallel to time axis, so that OC = AD = t.
Then change in velocity in time t = BC - OA = BC - CD = BD
But BC = v and CD = OA = It
0
elt
---+time(s)
Hence, BD = v - u.
Lateral meristem transfonns stem of a plant into trunk of tree.
Q.8. (a) A spoonful of sugar is added to a beaker containing 500 mL of water and stirred
for a while. State any two observations
that you will make.
(b) Account for your observations.
(3)
Ans. (a) On dissolving sugar in water, we observe that
(i) sugar dissolves completely in water.
(ii) there is no change in the volume.
(b) The above observations
can be explained by saying that there is lot of empty space between
water molecules. Sugar molecules occupy these empty spaces and, therefore, there is no change in
volume of the solution.
.
According to velocity-time graph, the total distance covered by the object is obtained by the '.
area under the graph.
Hence, distance s travelled by the object = area of the trapezium OABC
= area of rectangle OADC + area of triangle ADB
1
= OA x OC + -(AD
2
x BD)
But OA = It, OC = t, AD= OC = t and BD = change in velocity = v - u = at, where a = uniform
acceleration of the object.
1
1
s = ut + - (t x at) = ut + - at2.
2
2
Q.ll. State the law of inertia. Why do we fall in the forward direction if a moving
stops suddenly and fall in the backward direction if it suddenly accelerates from rest?
bus
(3)
.-44
.: .' Science-IX
Ans. Law of inertia: An object is unable to change its state of rest or of uniform motion in
a straight line unless acted upon by an external unbalanced force.
When driver of a moving bus applies brakes suddenly, the bus slows down but our body tends
to continue in the same state of motion due to the inertia of motion. Consequently, we may fall in
forward direction or may collide with the panels in front.
When driver of a bus accelerates it from rest, the bus gains speed but our body tends to remain
in its state of rest due to the inertia of rest. Consequently, we may fall backward or experience a
backward push.
Q.12. What happens to the
(i) distance between the
(ii) mass of both objects
(iii) mass of both objects
and
magnitude of the force of gravitation between two objects if :
o,bjects is tripled?
doubled?
as well as the distance between them is doubled?
(3)
Ans. We know that magnitude of the gravitational
1112separated
by a distance d is given by
F = G mld ~2
(i) When the distance
Force F' = G.
force between
two objects of masses
1111
However, units of force are selected in such a way that if III = 1 and a = 1, then force F also
becomes 1. In such a case
1 = k x 1 x 1 or k = 1
Therefore, the relation becomes F = l11a
One newton force is the force acting on an object of mass 1 kg
51 unit of force is 1 newton.
p'.'Oduces an acceleration of 1 m/s2 in it.
Q.14. A stone dropped from a window reaches the ground in 0.5 seconds:
(i) Calculate its speed just before it hits the ground.
(ii) What is its average speed during 0.5 s ?
(iii) Calculate the height of window from the ground.
(3)
Ans. As per question initial speed of stone u = 0, acceleration of stone = acceleration due to
gravity g = 9.8 m/s2 (in downward direction), and time taken by the stone to reach the ground
t = 0.5 s.
(i) Final speed of stone just before it hits the ground
v = u + gt=O + 9.8 x 0.5=4.9 m/s.
where G is a constant
,
between
d,2
(ii) The average speed of stone
the objects is tripled,
11111112=
G 11111112=
(3d)2
G1I111112=
9d2
d'
=
u +v
3d
0 + 4.9
= Vaverage = -2-
£
111'1 = 2//11
and
= -2-
= 2.45 m/ s.
~#,we
9
(ii) When mass of both the objects is doubled,
A-45
CSSE CCE Sample Question Paper
(iil) If height of window from the ground be h m, then using the relation h = ul + 2
111'2 = 2m2
have
Force F' = --Gmim2
= ----G·(2ml)(2m2)
d2
(iii)
-
d2
4G1I1111l2
d2
=4 F
When mass of both objects as well as the distance between
= 2d.
1
them is doubled,
h = 0 + - x 9.8 x (0.5)2 = 1.225 m.
2
we have
Q.15.
State two ways in which phloem is functionally different from xylem.
Draw a neat diagram of a section of phloem and label four parts.
Ans. (a) (I) Phloem unlike xylem allows movement of material in both the directions.
(ii) It transports food from leaves to other parts of the plants.
111'1 = 21111, 111'2 = 21112 and d'
(n)
(b)
Force F' =
Gmim2
d,2
=
G·(2ml)(2m2)
(2d)2
=
=F
4Gml1112
d2
Q.13. Derive the relation between force and acceleration. Define one unit of force.
(3)
Ans. Let an object of mass m was initially moving with a velocity u. It means that its initial
momentum PI = mu.
(b)
Sieve plate
If on applying a constant force F for time I the velocity of the object changes to v, then final
momentum of given object P2 = mv.
:..
Change in momentum = P2 - PI = mv - mu = m(v - u)
\
Rate of change of momentum
h
were
a = --v-u
t
l'
= acce eration
P2 - PI
= ---
I
m(v - u)
= ---
I
Phloem
= ma,
parenchyma
Companion
According to statement of second law of motion, the rate of change of momentum
proportional to the applied force. Hence, we have
=>
Sieve tube
0 f t h'e gIven 0b'
ject.
F oc
I
P2 - PI
or
F oc
(3)
ma
F = kma, where k is a constant of proportionality.
cell
is directly
Section of phloem
Q.16. Give one important functional difference amongst the muscle tissues and draw a
labelled diagram of the muscle tissue which never shows fatigue.
(3)
, Science-IX
A-46
Ans.
Distinction between
Striated
muscles
striated muscles, smooth muscles and cardiac muscles
Cardiac
Smooth
muscles
Do help
not
work
movement.
Example:
i.e.,
invollmtary
involuntary.
according
toExample
will
"
they
heart
to
They
are
also
parts
of
body.
pump
blood
to
all
:
mixture
do keeping.
notthrough
settle
down
small.
Type
of
paper
because
Filterability
the
particle
pass
size
is
a very
filter
on
system.
Its
particles
Stability
The labelled diagram of the muscle tissue which never shows fatigue is shown below:
Striations
A-
eSSE eeE Sample Question Paper
Q.20. (a) Distinguish among true solution, suspension and colloid in a tabular form under
the following heads :
(i) Stability
(ii) Filterability
(iii) Type of mixture
(b) What is meant by concentration
of a solution? How will you prepare a 10% solution
of glucose in water?
(5)
Ans. (a) Distinction behveen "true solution, suspension and colloid considering the factors
stability, filterability and type of mixture is gi\'en below in the tabular form:
Solutions
isare
Solution
is
Colloid
asmall.
stable
is
atime.
stable
SO/Iltioll
Colloid
after
homogeneous.
Its
particles
do
but
heterogeneous.
appear
homogeneous.
are
The
soluteSuspensions
size
because
the
do
through
Colloidal
particles
not
filter
issome
large.
pass
the
the
paper
particles
through
particle
filter
pass
sizenot
particles
settle
down
settle
on
down
keeping.
Suspensioll
Suspended
particles
asystem.
stable
system.
Itspaper
Suspension
is
not
Colloids
are
hete~ogeneous
Property
Nuclei
Cardiac muscles
Q.17. Which cell organelle would you associate with elimination
of old and worn out
cells? Why?
(3)
Ans. Lysosomes. They are capable of breaking down all organic material and keep the cell
dear by digesting worn out cell organelles. They are membrane bound sacs filled with powerful
,iigestive enzymes. When the cell is worn out and needs to be destroyed, the Iysosomes burst and
the enzymes digest the cell.
Q.18. State one difference
between dugwells
and tubewells.
Explain any two fresh
initiatives taken to increase the water available for agriculture.
(3)
Ans. Dugwells : In a dugwell, water is collected from water bearing strata.
Tubewells : Tubewells can tap water from deeper strata. From these wells, water is lifted by
pumps for irrigation.
Fresh initiatives for increasing the water available for agriculture
(i) Rainwater harvesting,
and
(ii) Water shed management.
includes:
.
This involves building small check dams which lead to an increase in ground water level. The
check dams stop the rainwater from flowing away and also reduce soil erosion.
Q.19. (a) Which two factors bring about loss of food grains during storage ? Give one
example for each.
(b) State any two control measures
to be taken before grains are stored.
(3)
Ans. (a) Factors which bring about loss of food during storage :
(i) Biotic (living) e.g., insects/rodents/fungi/mites.
(ii) Abiotic (non-living) e.g., inappropriate
temperature/humidity
at the place of storage.
(b) (i) Strict cleaning of the product before storage.
(ii) Proper drying of the product first in sunlight and then in shade and fumigation using
chemicals that can kill pests.
(b) Concentration
of solution is defined as the amount of the solute present in a given
amount (mass or volume) of the solution (or solvent).
A 10 per cent solution (by mass) of glucose can be prepared by dissolving 10 g of glucose in
90 g of water, so that the total mass of the solution is 100 g.
Or
Draw a neat and labelled diagram of the apparatus used to separate components
ink. Name the process and state the principle involved.
(b) Identify the physical and chemical changes from the following:
(i) Burning of magnesium in air.
(ii) Tarnishing of silver spoon.
(iii) Sublimation
of iodine.
(iv) Electrolysis of water.
Ans. (a)
Glass rod
(a)
of
blue-black
Paper clips
Jar
Stripof filter
paper
Stripof filter
paper
Linedrawn
pencil
by
Spot of ink
(a)
(b)
The process involved in the separation of components of blue-black ink is called chromatography,
-48
.. , Science-IX
Principle: The coloured component thatattraction
is moreof soluble in water rises faster and in this way the
forces
of particles
spaces
Kinetic
energy
Inter-particle
constant.
kineticInter-particle
unchanged.
keep
theenergy.
shape
colours of different dyes present in blue black ink get separated. compressible.
(b) Physical and chemical changes:
(i) Burning of magnesium in air
Chemical change
(ii) Tarnishing of silver spoon
Chemical change
(hi) Sublimation of iodine
Physical change
(iv) Electrolysis of water
Chemical change
Q.21. (a) State one similarity and one difference between evaporation and boiling.
(b) List four factors which affect the rate of evaporation.
(5)
(c) Describe
an activity to show that water vapour is present in air.
The liquid state changes into the gaseous state.
Ans. (a) Similarity:
Difference: Evaporation can take place at any temperature while boiling takes place at a fixed
temperature. For example, evaporation of water can take place at any temperature while boiling of
water takes place at 100°C only.
(b) Factors that affect the rate of evaporation:
The following factors affect the rate of
evaporation:
(i) Surface area· of the liquid exposed to air.
(ii) Temperature of the liquid.
(iii) Humidity of the surrounding air.
(iv) Wmd velocity.
(c) Activity to show that water vapour is present in air:
CSSE CCE Sample Question Paper
Compressibility
particles of a gas.
:r-
Water drops
Mathematically, momentum of an object is given by the product of its mass and velocity.
:.
Momentum (P) = mass (m) x velocity (v)
(i) Take about 200 g crushed
ice in a tumbler.
(ii) Keep it in the open for about 15 minutes.
We shall find that water droplets are seen on the outside surface of the tumbler.
This is because the water vapours in the air comes into contact with the cold surface of the
tumbler and are condensed to water droplets.
Or
Characteristic
Solids
Gases
SolidsLiquids
Gases
Liquids
are they
rigid.
are
are
completely
fluids,
fluids.
expand
easily.
i.e.,
They
to
can
any
can
flow
volume.
has both magnitude
and direction
i.e., it is a vector. S1 unit of momentum
is
~timeItt is
= 8given
s. that mass of object In = 50 kg, initial velocity u = 4 m s-l, final velocity v = 8 m s-I and
ice
Distinguish solids, liquids and gases in a tabular form under the following
(i) Rigidity
(ii) Compressibility
(iii) Inter-particle
forces of attraction
(iv) Inter-particle
spaces
(v) Kinetic energy of particles.
Ans. Distinction between soUds, liquids and gases covering the required
below in tabular fonn .
State its SI unit.
An object of mass 50 kg is accelerated uniformly from a velocity of 4 m S-1 to 8 m s-1 in 8 s.
Calculate the initial and final momentum of the object. Also find the magnitude of the force
exerted on the object.
.'
(5)
Ans. The total quantity of motion contained
in an object is called its momentum.
1kgms-l.
Crushed
attractions
and
Solids
are
not
directions.
between
the
energy.
between
the
particles.
remains
force
of
attraction.
kinetic
Solid volume
particles
Negligible
space
inter-particle
They
have
strong
space
is
there.
possess
very
low
inter-particle
possess
energy.
high
The
particles
can
They
have
weaker
can
bespace
Particles
aextremely
gas
Some
inter-particle
move
freely
inkinetic
all
attractions.
might
change
The
!Jut
shape
the
week
inter-particle
compressed
slightly.
in
liquid
Very
large
Highly
compressible.
Gases
have
Q.22. Define momentum.
Momentum
3
A-49
characteristics:
:. Initial momentum
PI
Final momentum
of object
= mu = 50 x 4 = 200 kg m s-I
of object
P2 =
"!v = 50 x 8 = 400 kg m S-I
Force exerted on the object
F=
P2 -
t
PI = 400 - 200 = 25 N.
8
Or
State the law of conservation of momentum. Why is a person hit harder when he falls on a
hard floor than when he falls on sand from the same height?
A bullet of mass 20 g is fired horizontally with a velocity 100 m s-1 from a pistol of mass
1.5 kg. Calculate the recoil velocity of the pistol.
points
is given
Ans. According to the principle of conservation of momentum, for an isolated system (i.e.,
when no external unbalanced force is acting on the system) the total momentum of the system
remains conserved. It is a basic conservation law of nature and is always true (or it never fails).
As an illustration if two objects moving on a smooth floor collides, then the sum of momenta of
the two objects before collision is equal to the sum of their momenta after collision.
.
When a person falls from a height on a hard concrete floor, he immediately comes in rest
position. It means change in momentum is taking place in an extremely short time and consequently,
, Science-IX
50
)fce exerted by the floor on the person to destroy its momentum is extremely large. Hence, chances
f more injuries. When a person falls ,?n a sandy surface, the surface gets compressed downward
nd it increases the time of fall. As a result for same change in momentum force exerted by sandy
urface on the person is less and chances of his being hurt are less.
As per question mass of bullet
= 20 g = 0.02 kg, muzzle velocity of bullet v = 100 In s-1, and the
:lass of pistol M = 1.5 kg
1/1I'
0.02 x 100
1
:. Recoil velocity of the pistol V = - -M = - ---- 1.5
= - 1.33 m s-
A-\.
eSSE eeE Sample Question Paper
I
Distance covered 5 = .!.[15 x 30] = 225 m.
2
Or
The velocity-time
graph of a body is given as follows:
111
The negative sign indicates that the direction of recoil of pistol is opposite
notion of bullet.
Q.23. The velocity-time
graph for an object is shown in the following
~
~.s 20 I
t
to the direction of
figure.
C
o
50
10
20
1, 40
(ii)
.s
(iil)
>
(iv)
~ 30
-0
o
Ans.
uniform
~
20
10
40
State the kind of motion represented by OA; AB.
What is the velocity of the body after 10 s and after 40 s ?
Calculate the retardation of the body.
Calculate the distance covered by the body between 10th and 30th second.
(i) The graph line OA represents uniformly accelerated motion but the line AB represents
linear motion.
(ii) Velocity of the body after 10 s is
20 m s-1 and after 40 s the velocity is zero.
~In 40
The motion is a retarded
motion
between 30 sand 40 s. In this duration, velocity
~
~ 20
~-~
Retardation
a = --falls from 20
m ~-1 to o.
>t
(iii)
o
5
10
15
_Timet(s)
20
of motion that the above graph represents.
(ii) What does the slope of the graph represent?
(iil) What does the area under the graph represent ?
(iv) Calculate the distance travelled by the object in 15 s.
Ans. (i) As velocity-time graph is a straight
50
line inclined to the axes, it represents a uniformly
:l.ccelerated motion.
(ii) The slope of the graph represents· the
value of uniform acceleration of the object.
(iii) . The area under
the graph represents
the distance covered by the object d':lring a given
time.
(iv) From the graph distance travelled by
the object in 15 s = area under velocity-time graph
= area of triangle OAB
1
= -[OB
2
t2 - t1
=-2ms-2
(5)
1, 40
.s
g
~
x BA]
From the graph OB = 15 sand BA = OC = 30 m s-1
O-W
= --40 - 30
B
A
Lh:
II E
0
10
20
_
30
+-c
40
Time(s)
(iv) Distance covered by the body between 10th and 30th second
= Area of the rectangle EABD = EA x ED
= (20 m s-l) x (30 s -10 s) = 20 x 20 = 400 m.
Q.24. How can crop variety improvement methods come to the rescue of fanners facing
repeated crop failures ? Describe three factors for which they could do crop improvement.
Which is the most common method of obtaining improved variety of crops? Explain briefly.
(5)
~ 30
t
.s
25
(i) State the kind
5
30
_Time(s)
(i)
t
B
A
~\n 40
~
20
10
---rB
0
5
10
15
_Time(s)
20
25
Ans. Three factors for which farmers could do crop improvement :
(i) Biotic and abiotic resistances:
Crop production can go down due to biotic (diseases,
insects and nematodes) and abiotic (drought, salinity, water logging, heat, cold and frost), stresses
under different situations. Varieties resistant to these stresses can improve crop production.
(ii) Change in maturity duration : The shorter the duration of the crop from sowing to
harvesting, the more economical is the variety.-Such short durations allow farmers to grow multiple
rounds of crops in a year. Uniform maturity makes the harvesting process easy and reduce~ losses
during harvesting.
,.
U-Like Science-IX
; 52
Wider adaptability:
Developing varieties for wider adaptability will help in stabilising
the crop production under different envirorunental conditions. One variety can then be grown
under different climatic conditions in different areas.
Most corrunon method is Hybridisation which involves crossing two varities having genes for
desired characteristics and bringing them together into a new \'ariety called hybrid.
Or
(iii)
A poultry farmer wants to increase his broiler production.
Explain three
practices he must follow to enhance the yield.
In what way is the daily food requirement of broilers different from those of
Ans. (a) Management practices the farmer must follow to enhance the yield
(i) Maintenance
of temperature:
To much high or low temperature affects
broilers.
(ii)
(iii)
(b)
management
egg layers?
are:
the growth of
Provision of hygienic conditions in housing and broiler feed.
Prevention and control of diseases and pests.
Broiler chickens must be fed with protein rich with adequate fat, vitamins A and K diets
for good growth
B
Q.25. Four students prepared mixtures in wat~r by taking sugar, sand, chalk powder and
starch respectively, in four different test tubes. After stirring, the mixture that appeared clear
and transparent was that of
(1)
(a) starch and water.
(b) chalk powder and water.
(c) sand and water.
(d) sugar and water.
Ans. (d) Out of sugar, sand, chalk powder and starch, only sugar is soluble in water.
Q.26. Rohit mixed starch with water, boiled the mixture well and stirred it. He observed
that
(l)
starch floats on the surface of water.
starch settles down at the bottom.
(c) starch forms a translucent mixture.
(d) starch forms a transparent
mixture.
Ans. (c) Starch forms a colloidal solu?on with water, which is translucent.
Q.27. You are provided with a mixture of iron filings and sulphur powder. When you add
carbon disulphide
to the mixture, you would observe
(1)
(a) iron particles dissolve and the solution turns black.
(b) sulphur
powder dissolves and the solution turns colourless.
(c) sulphur powder dissolves and the solution turns yellow.
(d) iron particles dissolve and the solution turns grey.
Ans. (c) Iron particles do not dissolve in carbon disulphide. Sulphur dissolves in carbon
disulphide forming a yellow solution.
Q.~8. A strip of magnesium metal is burnt in the flame. It is observed that
(1)
(a) a yellow light appears.
(b) a white dazzling light appears.
(a)
(b)
A-53
eeE Sample Question Paper
(c) magnesium starts melting.
lot of black smoke is produced.
Ans. (b) Magnesium metal burns in the flame with a white dazzling light.
Q.29. For determining the melting point of ice, the thermometer should be kept
(l)
(a) with its bulb in the ice cubes.
(b) in contact with the inner wall of the beaker.
(c) a little above the ice cubes.
(d) in touch with the beaker from outside.
Ans. (a) A correct melting point will be obtained only if the thermometer is kept with its
bulbs in the ice cubes. In any .other conditions (b), (c) or (d), the temperature obtained will be
different from melting point.
Q.30. A student takes some water in a beaker· and heats it over a flame for determining its
boiling point. He keeps on taking its temperature reading. He observes that the temperature
(1)
of the water
(d)
keeps on increasing regularly.
keeps on increasing irregularly.
(c) first increases slowly, then decreases rapidly and eventually becomes constant.
(d) first increases gradually and then becomes constant.
Ans. (d) The temperature keeps increasing till the boiling points is reached. At the boiling
point, the temperature becomes constant.
Q.31. The colour of sodium chloride 'and ammonium chloride respectively is
(1)
(a) yellow and white.
(b) white and yellow.
(c) both are white.
(d)' grey and yellow.
Ans. (c) Both sodium chloride and arrunonium chloride are white coloured compounds.
Q.32. In the laboratory, carbon disulphide is used as a solvent to separate a mixure of iron
filings and sulphur powder. What precaution has to be taken with carbon disulphide 7 (1)
(a) Keep away from water.
(b) Keep away from flame.
(c) Keep away from air.
(d) Keep away from iron sulphide.
Ans. (b) Carbon disulphide is an inflarrunable compound. Therefore, it has to be kept away
from flame.
(a)
(b)
rate and better feed efficiency.
SECTION
eSSE
Q.33. When iron nails are placed in copper sulphate solution,
colour disappears and the solution appears
(a) reddish brown
(b) blue
(c) light blue
(d) greenish
Ans. (d) The following reaction takes place
Fe + CuS04 ~
FeS04 + Cu
Ferrous
Sulphate
(green colour)
after 10 minutes,
its blue
(1)
Science-IX
A-54
Q.34. In an experiment to separate the. components of a mixture of sand, common
ammonium chloride, the component which will be removed by filtration is
salt and
(1)
sand
(b) common salt
(c) ammonium
chloride
(d) none of these
Ans. (n) In fact anunonium chloride is first remQ\'ed by sublimation. Then we dissolve the
mixture in water. Conunon salt dissolves leaving behind sand which is separated by filtration.
Q.35. To study the third law of motion, following sets of apparatus
are available in a
(1)
laboratory :
Set (i) One spring balance, two weight boxes, inextensible thread, one pulley with a clamp,
two pans of known mass.
Set (ii) Two identical spring balances, one weight box, inextensible thread, one frictionless
pulley with a clamp, one pan of known mass, a rigid support.
Set (iii) Four identical spring balances, two pulleys, inextensible thread, two clamps, two pans
of known masses, two rigid support.
(a)
Set (iv) Two identical spring balances, two weight boxes, two rigid supports, two pans of
known masses, inextensible thread, two frictionless pulleys with clamps.
To perform the experiment
successfully by using minimum apparatus,
the best choice
would be:
(a) Set (i)
(b) Set (il)
(c) Set (iii)
(d) Set (iv)
Ans. (b) To perform the experiment successfully by using minimum apparatus, the best
choice would be Set (ii) in wh.ich we use two identical spring balances, one weight box, inextensible
thread, one frictionless pulley with a clamp, one pan of known mass and a rigid support.
Q.36. For doing the experiment,
"to study the third law of motion using two spring
(1)
balances", four students A, B, C and D set up their apparatus as shown below:
----
----
'e\ Pulley
Pan
A,
eSSE eeE ~ample Question Paper
Ans. (a) The best set up is that of student A because in this set up both the spring balance1>
are set exactly in horizontal position.
Q.37. The appearance of magenta colour, on adding cone. Hel to a given sample of solution
of dal confirms the presence of
(1)
argemone oil in the dal.
potassium dichromate il1'the dal.
(c) saw dust in the dal.
(d) metanil yellow in the dal.
Ans. (d) Metanil yellow on reacting with cone. HCl gi\'es magenta colour.
Q.38. The steps for conducting the starch test on the given sample of rice grains are (l)
(i) crush the rice grains.
(ii) add water to the test tube.
(iii) add few drops of iodine.
(iv) boil the contents and filter.
The most appropriate order in which the steps should followed are:
(n)
(b)
W~~~~
W~~~~
(c) (iii), (iv), (I), (il)
(d) (I), (ii), (iv), (iii)
Ans.
obtained.
Q.39.
the cells
(d) After crushing rice, water is added to it and then after boiling it, starch solution is
Lastly iodine solution is added to detect the presence of starch.
While preparing a temporary mount of the cheek cells, the reason behind staining
(1)
is
prevent the cells from drying quickly.
preserve them.
(c)
disinfect them.
(d)
make the organelles clearly visible.
Ans. (d) Cell organelles when stained become more visible.
Q.40. Which of the observations noted by Arun about the parenchyma
(a)
(b)
to
to
to
to
(l)
(A)
The cells are are thin walled.
(b) Large cells are placed together with intercellular spaces.
(c) The cells are loosely packed.
(d) The cells are thick walled.
Ans. (d) Characters given in (a), (b) and (c) are identifying features of parenchyma tissue.
Q.41. The formula used to calculate the percentage of water absorbed by raisins is
(1)
(a)
W-W
2__
1
x100
WI
Pan
W2 in the formula refers to :
(B)
The best set up is that of
(a) student A
(c) student C
mass of
mass of
(c) mass of
(d) mass of
Ans. (b) This
(a)
Pan
Pan
tissue is not correct?
(b)
(0)
(C)
(b)
student B
(d) student D
raisins before absorption of water.
raisins after absorption of water.
water left in the beaker at the end.
water absorbed by the raisins.
is the correct answer.
Science-IX
Q.42. One of the following shows the correct diagrammatic representation of a striated
(1)
muscle fibre when seen under the low power of a microscope ?
(A)
(B)
(C)
(D)
The correct answer is
(a) A
(b)
C
(d)
B
D
Ans. (a) In striated muscles their are more than one nuclei located towards the periphery of
cell and the striation is alternate.
(c)