class 9 science chapter gravitation pdf

GRAVITATION
3.1 INTRODUCTION
Any two objects in this universe attract each other. This property is
known as gravitation and the force responsible for the attraction is
called gravitational force.
Question based on basic knowledge required to understand
this chapter
1. Aparticle moving with a velocity of 10 m/s is moving with a uniform
acceleration of 5 m/s2
. What will be the final velocity of the particle
after 5 s:
(A) 30 m/s (B) 32 m/s (C) 36 m/s (D)35 m/s
2. A particle accelerates from rest with an acceleartion of 10 m/s2
and then decelerates at the rate of 15 m/s2
. If the particle was in
motion for a total time of 10 s then find the maximumvelocity of the
particle.
(A) 50 m/s (B) 60 m/s (C) 20 m/s (D)25 m/s
3. In the above question i.e. Q.2 the distance travelled by the particle
in 10 s will be:
(A) 200 m (B) 250 m (C) 300 m (D)310 m
4. A bullet going with speed of 175 m/s enters a concrete wall and
penetrates a distance of 2.5 cm before coming to rest. Find the
deceleration:
(A) 61250 m/s2
(B) 62150 m/s2
(C) 61000 m/s2
(D) 60100 m/s2
5. A particles moving with a speed of 10 m/s along the +X direction.
The particle is having an acceleration of 5 m/s2
along –X direction.
The particle comes to rest after a time of:
(A) 1 s (B) 2 s (C) –2 s (D) 3 s
6. A quantity has a value of –6 m/s. It may be the
(A) speed of a particle (B) velocity of a particle
(C) acceleration of a particle (D) postition of a particle
7. A train accelerates from 20 km/h to 80 km/h in 4 minutes. The
distance covered by train during this period is (Assume the tracks
are straight)
(A) 10 km (B)
3
10
km (C)
4
10
km (D)
9
10
km
3.1 Introduction
3.2 Universal Law of
Gravitation
3.3 Free Fall
3.4 Relation between
acceleration due to
earth’s gravity g and
gravitational constant G
3.10 Mass and Weight
3.11 Weightlessness
3.12 Thrust & Pressure
3.13 Byoyancy
3.14 Archimedes Principal
3.15 Floating or Sinking of an
object in a fluid
3.16 Density and Specific
Density
“IIT-JEE Foundation”
*3.5 Variation in the value of
acceleration due to
gravity
*3.6 Orbital Speed of Satellite
*3.7 Time Period
*3.8 Escape Velocity
*3.9 Motion of Planets
VAVA CLASSES/PHY/9TH
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8. A girl bends to touch her toes. Then the motion of head is an example of:
(A) UniformMotion (B) Accelerated Motion
(C) Constant Velocity Motion (D) Simple Harmonic Motion
9. Aforce F acting on a 2 kg body produces an acceleration of 1.5 m/s2
. Another force F2 acting on a 5.0 kg
body produces an acceleration of 2.0 m/s2
. The ratio F2/F1 is:
(A) 2 (B) 2.5 (C) 3 (D) 1
10. A force produces an acceleration of 5.0 cm/s2
. when it acts on a body of mass 20 g. Then the force in
newtons is:
(A) 10–4
N (B) 10–2
N (C) 10–3
N (D) 10+5
N
3.2 UNIVERSAL LAW OF GRAVITATION
Every object in the universe attracts each other.According to newtons law of gravitation attraction force
between two bodies is directly proportional. To the product of their masses and inversely proportional to
the square of the distance between them. The direction of force is along the line joining the centre of two
bodies.
Let two object A and B of masses m1
, m2
are seprated by the distance d then attraction force between
them is
1 2
F m m

2
1
F
d

1 2
2
m m
F
d
 

 
 
1 2
2
Gm m
F
d

Where G is proportionality constant and is called the universal gravitational constant its value is.
2
2
11
10
67
.
6
kg
m
N
G


 
Illustration 1
Two bodies A and B of masses m and 4m respectively are kept at a distance 'd' from each other.
Where should a particle of mass m/2 be placed between the line joining the bodies A and B so that
net force on it is zero.
(A) d/4 (B) –d (C) d/3 (D) 2d
Solution
2
x
2
m
Gm 





=
2
x)
(d
2
m
G(4m)







m m/2
x (d-x) 4m
2
x
x
–
d






= 4 
x
x
–
d
= ±2
x
x
–
d
= 2 d = 3x

x =
3
d
This point lies in between m & 4m
x
x
–
d
= –2
d – x = –2x
d = –x
x = –d This is outside point, so discarded.
 (C)
3.3 FREE FALL
Common example of motion along straight line with constant acceleration is free fall of a body near the
earth surface. If air resistance is neglected and body is dropped near the surface it falls along a vertical
line. The acceleretion is in the vertically downwards direction and its magnitude is almost constant If
height is small as compare to the radius of earth The magnitude is approximately equal to 9.8 m/s2
denoted by ‘g’. The equations of motion during free fall are as under
v = u + gt , S = ut + 1
/2
gt2
and v2
= u2
+ 2gh.
by convection we take ‘g’ as negative in upward direction and positive in downward direction so during
down ward motion, we use
v = u–gt
S = ut –
2
1
gt2
; v2
= u2
–2ah
3.4 RELATION BETWEEN ACCELERATION DUE TO EARTH’S GRAVITY
g AND GRAVITATIONAL CONSTANT ‘G’
Let mass of the earth be Me
and its radius. be Re
and the whole mass Me
is concentrated at its centre. Let
a body of mass m be situated at the surface of the earth at a small height above. The surface distance, of
the body from the centre of earth may be taken as Re
. according to the law of gravitation, the force of
attraction acting on the body due to the earth is given by
2
e
e
R
m
GM
F 
According to Newton second law of motion
 
mg
F   2
e
e
R
m
GM
mg 
2
e
e
R
GM
g 
In terms of density g =










2
3
3
4
e
e
R
R
G 

= 





g

 Re
3
4

*3.5 VARIATION IN VALUE OF ACCELERATION DUE TO GRAVITY
The value of acceleretion due to gravity g varies from place to place at the surface of the earth and also
we go above or below the surface of the earth.
(i) Due to shape of earth: Earth is not perfectly a sphere but some what flat at the two poles.
Its equtorial radius is nearly 21 km longer than polar radius we know the value of g depend upon the radius
of the earth.






 2
e
e
R
GM
g  
e
p R
R 
Therefore 





 2
1
e
R
g Then  
e
p g
g 
(ii) The value of g decrease on going above the surface of earth
Let a body of mass m is placed above the h height on earth surface at that point acceleretion due to
gravity is g.
Then according to gravitaiton law. attraction force at that point
 2
h
R
m
GM
F
e
e
G

 ...(i)
Force due to Newton law F = Mg´ ...(ii)
Equation (i) and (ii) equal
 
e
2
e
GM m
mg'
R h


 2
'
h
R
GM
g
e
e

 2
e
e
R
GM
g 
 2
2
'
h
R
gR
g
e
e

  
2
e
e gR
GM 
2
' 







h
R
R
g
g
e
e
h
R + h
e
Re
2
'






 

e
e
R
h
R
g
g
2
1
'









e
R
h
g
g








R
h
g
g
2
1
' {using binomial expansion. (1+x)n
=1+nx}
(iii) The value of g decrease on going below the surface of earth: When we go in the (d) depth of
earth the value of g’ will be

on earth surface
3
3
4
e
e
R
M

 
g
g 
 Re
3
4

A
B
O
d
at depth   g
d
R
g e 
 

3
4
1
Dividing the both equation
 
e
e
R
G
g
d
R




3
4
3
4

e
e
R
d
R
g
g 

'










R
d
R
R
g
g
e
e
'








R
d
g
g 1
' 







R
d
g
g 1
'
Illustration 1
At what height abouve the earth’s surface the acceleration due to gravity will be 1/4th of its value
at the earth surface redius of earth is 6400 km.
Solution
2
1
'










e
R
h
g
g
2
1
9










e
R
h
g
g
3
1 


e
R
h
h = 2Re 2800 km
Illustration 2
The depth d at which acceleration due to gravity become 1/4 time the value at the surface.
Solution
 
R
d
g
g 
 1
'   
g d
g 1
R
4
 
R
d

1
4
1

4
1
1 

R
d
4
3
4
1
1 


R
d
4
3R
d 

*3.6 ORBITAL SPEED OF SATELLITE
When a saitelite provided by revolves in a circular orbit around the earth, a centripetal force acts upon the
satellite this force is the gravitational force exerted by the earth on the satellite a satellite of mass m is
revolving around the earth with a speed V0 in a circular orbit of radius r.
The centripetal force on the satellite is
2
o
e
mV
F
r
 

 
 
Let Me be the mass of earth the gravitational force exerted bythe earth on the satellite will be
e
e 2
GM m
F
r
 

 
 
As the gravitational force provide the required centripetal force the
Fe = Fc
r
mV
r
m
GM o
e
2
2
 n
R
r
GM
V e

0 [ r = Re + h]
Where Re = radius of earth
h = height from the surface of earth
*3.7 TIME PERIOD
Time taken by the satellite to complete one revolution around the planet is called time period.







0
2
V
r
T
 3
e
r
T 2
GM
 
 
 
 
 
*3.8 ESCAPE VELOCITY
The minimum velocity of the body by which if the body is thrown up, it will never return to the earth, this
minimum velocity of the body is called the escape velocity.
The gravitational potential energy of a body of mass m placed on the earth’s surface is given by








e
e
R
m
GM
U
and required K.E [KE =
2
2
1
e
mV ]
By conservation of energy total energy at earth is equal to total energy at 
0
0
2
1 2



 e
e
e
mV
R
m
GM
e
e
e
R
GM
V
2


*3.9 MOTION OF PLANETS
There are certain celestial bodies which revolve around the planets these are called satellites for example
moon, revolve around the earth similarly mars has two satelites.
Kepler’s laws: Kepler found important regularities in the motion of planets these regularities are known
as Kepler’s three laws of planetry motions.
(i) Kepler’s First law: (Orbit’s law):All planets move around the sun in elliptical orbits, having the sun
at one focus of the orbit.
(ii) Second law: (Area’s law): A line joining any planet to the sun sweeps out equal areas in equal time
interval that is the areal speed of the planet remains constant.
const
dt
dA

(iii) Third law : (Periods law): The square of period of revolution (time of one complete revolution) of
any planet around the sun is directly propotional to the cube of the semimajor axis of its elliptical orbit.
 
3
2
r
T 
Illustration 3
The distance of two planets from the sun are 1013
m and 1012
m. find the ratio of time period .
Solution
3
2
r
T 
2
3
2
1
2
1









r
r
T
T
2
3
12
13
2
1
10
10









T
T 10
10
2
1

T
T
3.10 MASS AND WEIGHT
Now we are in better position to understand and define two quantities which we encounter frequently.
These quantities are mass and weight. We must be very clear that both are totally different. While
studying Newton's laws of motion. You must have studied about inertia. That is defined as "Tendency to
resist any change in state of motion of the body" You have studied that body of more mass requires more
force to change its state i.e. has more inertia. On the basis of these observation we define "mass as a
quantity responsible for inertia. It is one of the fundamental quantities and its unit is kilogram (kg).
Mass of the body does not changes place to place or plante to planet.
Now let us define the weight. Weight of the body on earth (Planet) is defined as "the gravitational force
of the earth (planet) on the body". So its a kind of force and its unit is Newton. Say a body is placed at the
surface of the earth. Its mass is m. Gravitational force of earth on the body.
F =
e
2
e
GM m
R
= weight of the body m earth
=
e
2
e
GM
m
R
 
 
 
 
F = mg = w

So weight of the body is mathematically equal to the product of mass and gravitational acceleration. Since
value of gravitational acceleration may different place to place and planet to planet so weight of the body
is different at different point.
Illustration 4
A body has a mass of 10 kg. The weight of the body on the earth and the moon is respectively: (take
gearth = 9.8 m/s and gmoon = 1.6 m/s2
)
(A) 98 N, 20 N (B) 98 N, 18 N (C) 98 N, 16 N (D) 100 N, 16 N
Solution
Wearth = mgearth = 10 × 9.8 = 98 N
Wmoon = mgmoon = 10 × 1.6 = 16 N
 (C)
Illustration 5
A body of mass 20 kg is weighed on the surface of the earth and then weighed at a height equal to
the radius of the earth. The difference in the weights will be:
(A) 120 N (B) 147 N (C) 139 N (D) 163 N
Solution
Wsurface = 20 × 9.8 = 196 N
Wheight 'h' =
4
8
.
9
0
2 
=
4
96
1
= 49 N
 (B)
3.11 WEIGHTLESSNESS
You must have seen astronuts in satellite. They float in the satellite. If we measure their weight in the
satellite, it comes out to be zero.
A body needs no support to stay at rest in the sattelite and hence all position are equally comfortable.
Water will not fall down from the glass even if it is inverted. Space in the satellite acts on gravity free hall.
Such a state is called weightlessness.
3.12 THRUST AND PRESSURE
In the previous chapter we have studied about force. Now let us take a common day today example.
When we push a blunt pin into the wood, it is hard to insert the pin in wood and requires high force. Now
if we push a sharp pin into the wood, the pin goes into the wood very easily with relatively very less force.
So we conclude that the effect of the force depends on the area of the object on which it acts.
We define thrust as "the force acting on a body perpendicular to its surface." Unit of thrust is same as per
that of force.
We now define one more quantity named as pressure. We define pressure as "Thrust acting on an unit
area of object."
P =
Thrust
Area
Unit of pressure is Pascal.
1 Pascal = 2
1m
1N
= 1N/m2
So if thrust acting on a surface of area 1m2
is 1N pressure acting on it its said to be 1 Pascal.

3.13 BUOYANCY
Now let us talk about fluid. In genral we define fluid as those substences which can flow easily.All liquids
and gases are fluids.Afluid exerts pressure in all the direction. So if a body is submerged partically or fully
fluid pressure is exested on the body. Since the fluid pressure is dependent upon depth in the fluid. So
different part of body feel different pressure, or different force, due to which a net force acts on body in
upward direction. So weight of the body in fluid appreans to be less. This property of fluid is called
Buoyancy.
The net upward force acting on the body due to fluid is called buayant force.
The buoyant force acting on the body depends upon three factors.
(i) Volume of the object immerced in fluid
(ii) Density of the fluid.
(iii) Acceleration due to gravity at the place where fluid is resting.
Greater volume of object means greater buoyant force on the object and greater density of fluid means
greater buoyant force on the body immersed. Buoyant force is not a random force It obeys a principle
given by archimedes called archimedes principle.
3.14 ARCHIMEDES PRINCIPLE
Archimedes principle gives us the value of Buoyant force.According toArchimedes principle "When an
object is wholly (or partially) immersed in a liquid, it experiences a buoyant force, which is equal to the
weight of liquid displaced by the object."
Buoyant force acting on an object = weight of the liquid displaced by the object
If a body of volume V is immersed (wholly) in a fluid of density  then the volume of the fluid displaced
by the body is also V and mass of the fluid displaced by body will be V.
So Buoyant force m body = Density × Displaced volume of fluid = V
3.15 FLOATING OR SINKING OF AN OBJECT IN A FLUID
When a body is immersed in a fluid, two forces act on it
(i) Its weight = mg
(ii) Buoyant force
If this Buoyant force is less than weight, object sinks and if Buoyant force is more than weight, the object
floats.
3.16 DENSITY & SPECIFIC DENSITY
Density of a body is defined as mass per unit volume say we have a body of mass m and volume V and
its density is  then
Density =
volume
mass
 =
V
M
Units of  is kg/m3
We define one more thing known as specific density, defined as
Specific density =
water
of
density
object
of
density
Specific density = o
/w
Note: If the density of object is less than the density of fluid it will float on fluid, but if it is greater than the
density of fluid it will since in the fluid.

EXERCISE-I
1. What is the value of universal gravitational constant?
2. A body has a weight of 10 kg on the surface of earth. What will be its weight when taken to the centre of
the earth.
3. What is the value of gravitational acceleration acting on a free falling object?
4. Mass of one object is 600 gram on the earth, what is its mass on the moon?
5. What is the value of gravitational acceleration ‘g’ on moon?
6. Weight of one person is 600 kg. wt. at earth. What is his weight on moon?
7. If distance between two objects halved, what happens to the gravitational force between them?
8. A stone is dropped from the top of a tower. Its velocity after it has fallen 20 m is (take = 10 m/s2
).
9. A body weighs 63 N on the surface of earth. What is the gravitational force on it due to the earth at a height
equal to half the radius of earth? Mass of earth is 6 × 1024
kg and Radius of earth is 6400 km,
G = 6.6 × 10–11
N-m2
/kg2
.
10. Assuming the earth to be a uniform sphere of radius 6400 km and density 5.5 g/cc, find the value of ‘g’ on
its surface, G = 6.67 × 10–11
N-m2
/kg2
.
EXERCISE-II
1. Two stones of masses 50 gram and 500 gram are dropped from a height. Which stone will reach the
surface of earth first and why? Explain your answer with the help of Newton’s law of gravitation?
2. Define gravitational acceleration and find its value at the surface of earth.
3. Write the relationship and differences between gravitational acceleration (g) and gravitational constant
(G).
4. Distinguish between mass and weight giving reasons.
5. Show that the weight of an object on moon is one - sixth of its weight on earth.
6. Explain how a projectile can be made to go round the earth.
7. How are ocean tides caused?
8. The earth attracts an apple. Does the apple also attracts the earth. If it does, why does the earth not move
towards the apple?
9. The mass and radius of a planet both are double of those of earth. What will be the mass of one man whose
mass is 60 kg?
10. Projectile is thrown horizontally from the top of a tower. It takes 2 s to hit the ground. Find the height of the
tower. Given g = 9.8 ms–2
.

EXERCISE-III
SECTION-A
 Multiple choice question with one correct answers
1. A geostationary satellite is orbiting the earth at a height of 6R above the surface of the earth, where R is the
radius of the earth. The time period of another satellite at a height of 2.5 R from the surface of the earth is:
(A) 2
5 hrs (B) 2
6 hrs (C) 2
3 hrs (D) 2
4 hrs
2. A particle is projected vertically upwards from the surface of earth (radius Re
) with a kinetic energy equal
to half of the minimum value needed for it to escape. The height to which it rises above the surface of earth
is:
(A)
2
R
(B)
3
R
(C)
3
2R
(D)
3
5R
3. The time period of an earth–satellite in circular orbit is independent of
(A) the mass of the satellite(B) radius of the orbit (C) none of them (D) both of them
4. The value of the acceleration due to gravity on a planet having three times the mass of the earth and radius
twice that of the earth:
(A) g
4
1
(B) g
4
3
(C)
3
g
(D)
5
g
5. The weight of an astronaut in a satellite revolving around the earth measured using a spring balance will be
(mass of astronaut = 60 kg)
(A) zero (B) 600 N (C) 60 N (D) 588 N
6. The height over the earth's surface at which the weight of a body become half of its value at the surface
willbe:
(A) R
2 (B) 2R (C)  R
1
2  (D)  R
1
3 
7. A particle of mass 50 g experiences a gravitational force of 2N when placed at a particular point. The
gravitational field at that point will be:
(A) 30 N/hg (B) 35 N/hg (C) 25 N/hg (D) 45 N/hg
8. If a cube of ice in water melts completely then the level of water
(A) remains unchanged (B) level falls down (C) level increase (D) none of these
9. The spring balance Areads 2 kg with a block m suspended from it. Abalance B reads 5 kg when a breaker
with liquid is put on the pan of the balance. The two balances are now so arranged that the having mass is
inside the liquid in the breaker as shown in the figure. In this situation:
(A) the balance A will read more tham 2 kg
(B) the balance B will read more than 5 kg
(C) the balance A will read more than 2 kg and B will read more than 5 kg
(D) the balance A and B will read 2 kg and 5 kg respectively
10. Aboat floating in a water tank is carreging a number of large stones. If the stones are unloaded into water,
what will happen to the water level:
(A) If falls (B) It rises up (C) remains unchanged (D) none of these

SECTION-B
 Multiple choice question with one or more than one correct answers
1. Time period of satellite revolving around a planet does not depend upon
(A) mass of the planet (B) orbital radius
(C) mass of the satellite (D) radius of the planet
2. A body is firstly placed at some height h from earth surface than taken at depth h below the earth
(h < < < R)
(A) Its mass changes
(B) Its mass increases, throwout
(C) Its mass increases, while journey up to earth surface and decreases when taken to depth
(D) its mass remain same.
3. Consider the previous situation again.
(A) weight of the body remain contant.
(B) weight of the body firstly increases up to the surface and then decreases
(C) weight of the body is maximum at surface and minimum at when its at height h (2n Minimum's
situation)
(D) weight of the body is maximum at h depth below the surface
4. A body of volume 8 m3
is placed gently at the surface of a fluid. Relative density of matter of body is 1/2.
(A) Body will float
(B) Its whole volume is at the surface of fluid on 1/4
(C) Its1/2 part its submerged in fluid
(D) it will since.
SECTION-C
 Comprehension
You are scintist at NASA in USA. you are involved in the desigen of a mission carrying humans to the
surface of the planet mans and then their retrun to the earth. Radius of mass is Rm
= 3.40 × 106
m and mass
Mm
= 6.42 × 1023
kg. The earth weight of the mass lander is 39200 N. You have to give a presentation
before other scientists and answer the questions asked here. (Take g at surface of earth 9.8 m/s2
and
neglect effect of all other planets.)
1. Mass of mass lander on mass, will be
(A) 4000 kg (B) 4500 kg (C) 5000 kg (D) 3200 kg
2. Acceleration due to gravity of mass will be at height 6 × 106
m above the surface of the mass
(A) 0.36 m/s2
(B) 0.40 m/s2
(C) 0.45 m/s2
(D) 0.48 m/s2
3. Weight of the mass lander on mass surface will be
(A) 16000 N (B) 14000 N (C) 12000 N (D) 15000 N
4. While returning back to the earth the minimum speed of mass lander required is
(A) 5 × 103
m/s (B) 8 × 103
m/s (C) 6 × 103
m/s (D) 2 × 103
m/s
SECTION-D
 Match the following (one to many)
Column-I and column-II contains four entries each. Entries of column-I are to be matched with some
entries of column-II. One or more than one entries of column-I may have the matching with the same entries

of column-II and one entry of column-II may have one or more than one matching with entries of column-I
1. Column I Column II
(A) Escape velocity (P) 2
R
Gm
(B) velocity of satellite (Q)
h
R
Gm

(C) gravitational acceleration (R) 2gR
(D) Time period (S)
3
(R h)
2
GM


here M  mass of the planet considered
R  Radius of the planet.
EXERCISE-IV
1. The mass of the earth is 6×1024
kg and that of the moon is 7.4×1022
kg. If the distance between the earth
and the moon is 3.84×105
km. calculate the force exerted by the earth on the moon. G = 6.7 × 10–11
m2
kg–
2.
2. A far falls off a ledge and drops to the ground in 0.5 s, Let g = 10 m s–2
(for simplifying the calculations).
(i) What is its speed on striking the ground?
(ii) What is its average speed during the 0.5 s?
(iii) How high is the ledge from the ground?
3. An object us thrown vertically upwards and rises to a height of 10 m. Calculate (i) the velocity with which
the object was thrown upwards and (ii) the time taken by the object to reach the highest point.
4. Mass of an object is 10 kg. What is its weight on the earth?
5. An object weighs 10 N when measured on the surface of the earth. What would be its weight when
measured on the surface of the moon?
6. A block of wood is kept on a tabletop. The mass of wooden block is 5 kg and its dimensions are 40 cm × 20
cm × 10 cm. Find the pressure exerted.
40
cm
40 cm
20 cm
20
cm
1
0
c
m
10
cm
( )
a ( )
b
7. Relative density of silver is 10.8. The density of water is 103
kg m–3
. What is the density of silver in SI unit?

EXERCISE-V
1. State the universal law of gravitation.
2. Write the formula to find the magnitude of the gravitational force between the earth and an object on the
surface of the earth.
3. What do you mean by free fall?
4. What do you mean by acceleration due to gravity?
5. What are the differences between the mass of an object and its weight?
6. Why is the weight of an object on the moon th
6
1
its weight on the earth?
7. Why is it difficult to hold a school bag having a strap made of a thin and strong string?
8. What do you mean by buoyancy?
9. Why does an object float or sink when placed on the surface of water?

Answers
Knowledge base questions
1. (D) 2. (B) 3. (C) 4. (A) 5. (B)
6. (B) 7. (B) 8. (B) 9. (A) 10. (C)
Exercise I
1. 6.67×10–11
Nm2
/kg2
2. 0 3. 9.8 m/sec2
4. 600 gm
5. 1.63 m/sec2
6. 100 kg-wt. 7. 4 times more 8. 20 m/sec
9. 28 N 10. 9.8 m/sec2
Exercise-III
Section-A
1. (B) 2. (B) 3. (A) 4. (B) 5. (A)
6. (C) 7. (D) 8. (A) 9. (B) 10. (A)
Section-B
1. (C,D) 2. (D) 3. (B,C) 4. (A,C)
Section-C
1. (A) 2. (D) 3. (D) 4. (A)
Section-D
1. (A)-(Q), (B)-(Q), (C)-(P), (D)-(S)

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