Courses

# Practice Questions, Gravitation, Class 9, Science | EduRev Notes

## Class 9 : Practice Questions, Gravitation, Class 9, Science | EduRev Notes

The document Practice Questions, Gravitation, Class 9, Science | EduRev Notes is a part of the Class 9 Course Science Class 9.
All you need of Class 9 at this link: Class 9

Q1. Let us find force of attraction between two block lying 1m apart. Let the mass of each block is 40 kg.
Sol. F= ?
m1 = 40 kg
m2 = 40 kg
d = 1 m Q2. The gravitational force between two object is 49 N. How much distance between these objects be decreased so that the force between them becomes double?

Sol. Step-1 : Here, F = 49 N
Let r = original distance between the objects.
m1 and m2 = masses of the objects
Gravitational force between these objects is given by

Step-2 : Now, F' = 2F = 2 × 49 N = 98 N
Let, r' = decreased distance between the objects
m1 and m2 = masses of the objects
Gravitational force between these objects is given by
Dividing (1) by (2), we get
[Q F' = 2F]

Q3. Two bodies A and B having masses 2 kg and 4 kg respectively are seperated by 2m. Where should a body of mass 1 kg be placed so that the gravitational force on this body due to bodies A and B is zero?

Sol. Let a body of mass 1 kg be placed at point P at a distance x from body A. Therefore, the distance of mass 1 kg from a body B = (2 x) m.
Gravitaional force on 1 kg due to body A is given by
F = N(along PA) = N (along PA)
Gravitational force on 1 kg due to body B is
F' = N (along PB).
Since gravitational force on 1 kg due to bodies A and B is zero, therefore,
F = F ' or (2 x)2 2x2 or (2 x) = x = 1.414 x
or 2.414 x = 2 or x = 0.83 m.

Q4. Calculate the force of gravitation due to a child of mass 25 kg on his fat mother of mass 75 kg if the distance between their centres is 1m from each other. Given G = (20/3) × 10-11 Nm2 kg-2.

Sol. Here m1 = 25 kg ; m2 = 75 kg ; d = 1 m ; G = × 10-11 Nm2 kg-2
Using F = 12,500 × 10-11
or F = 1.25 × 10-7 N

Q5. A boy drops a stone from the edge of the roof. If passes a window 2m high in 0.1s. How far is the roof above the top of the window ?
Sol. Let a stone be dropped from the edge of the roof A. Let it passes over B with a velocity say u. Consider motion BC.
u = ?, a = 9.8 ms_2 ; s = h = 2m ; t = 0.1 s
Using s = ut gt2, we have
2 = u(0.1) × 9.8 (0.1)2
2 = 0.1u 0.049
0.1u = 2 _ 0.049
or u = = 19.51 ms_1

This initial velocity at B in motion BC is the final velocity in motion AB. Considering motion AB, we have
u = 0 ; v = 19.51 ms_1 ; s = ? ; a = 9.8 ms_2
Using v2 _ u2 = 2as, we have
(19.51)2 _ (0)2 = 2 × 9.8 s
or s = 19.4 m
Roof is 19.4 m above the window.

Q6. A ball thrown up is caught by the thrower after 4 s. With what velocity was it thrown up? How high did it go? Where was it after 3 s ? (g = 9.8 m s_2)
Sol. Since the time of going up is the same as that of commg down, therefore time of going up = 4/2 = 2s. Let it starts upward with velocity u.
Here u = ?; a = - 9.8 m s_2 ; t = 2s; v = 0 (at the top); s = h
Using v = u at
or 0 = u _ 9.8 × 2

or u = 19.6 m s_1
Again v2 _ u2 = 2as
0 _ (19.6)2 = 2 (_9.8) h
h = 19.6 m
After 2s, it starts coming downwards (starting with u = 0). Considering downward motion.
u = 0; a = 9.8 m s_2; t = 3 _ 2 = 1s; s = ?
s = ut at2
or s = 0 × 9.8 (1)2 = 4.9 m from top.

Q7. Coconut is hanging on a tree at a height of 15 m from the ground. A boy launches a projectile vertically upwards with a velocity of 20 m s_1. After what time the projectile pass by coconut? Explain the two answers in this problem.
Sol. Here u = 20 m s_1; a = - 10 m s_2; s = 15 m ; t = ?
Using s = ut at2, we have
15 = 20t (_10) t2
Dividing throughout by 5, we have
3 = 4t - t2
or t2 _ 4t 3 = 0
or (t _ 1) (t _ 3) =t _ 1 = 0 or t = 1s
or t _ 3 = 0 or t = 3s
After 1s, it will cross coconut while going up and after 3 s while coming down.

Q
8. Fig. (a) shows how to calculate pressure exerted by a brick of mass 3 kg : (a) when standing on end; (b) when lying flat. The total force or thrust exerted is the same in both the cases.
Sol. In Fig. (a), A1 = 5 cm × 10 cm = 50 cm2 = 50 × 10_4 m2 and
F1 = 3 kg wt = 3 kg × 10 m/s2 = 30 N
Thus, pressure exerted, P1 = = 6 × 103 N/m2 = 6 × 103 Pa
In Fig. (b), A2 = 10 cm × 20 cm = 200 cm2 = 200 × 10_4 m2 and
F2 = 3 kg wt = 3 kg × 10 m/s2 = 30 N
Thus, pressure exerted, P2 = = 1·5 × 103 N/m2 = 1·5 × 103 Pa

Q9. A sealed tin of Coca-cola of 400 g has a volume of 300 cm3. Calculate the density of the tin.
Sol. Here, mass of tin m, = 400 g
Volume of tin, V = 300 cm3
Density of tin, r = 1.33 g cm3.

Q10. A sealed cane of mass 600 g has a volume of 500 cm3. Will this cane sink in water ? Density of water is 1 g cm3.
Sol. Here, mass of cane, m = 600 g
Volume of cane, V = 500 cm3
Density of cane, r = 1.2 g cm3.
Since, density of the cane is greater than the density of water, so the can will sink in water.

Q11. The density of water is 1000 kg m3. If density of gold is 19320 kg m3, find the relative density of the gold.
Sol. Density of water = 1000 kg m3
Density of gold = 19320 kg m3
Using, R.D. of gold = we get R.D. of gold

Q12. The density of water is 1000 kg m3. If relative density of iron is 7.874, then calculate the density of iron.
Sol. Density of water = 1000 kg m3
Relative density (R.D.) of iron = 7.874
Using, R.D. of iron = we get
Density of iron = R.D. of iron × density of water
= 7.874 × 1000 kg m3 = 7874 kg m3.

Q13. A plastic bottle of 500 g has a volume of 450 cm3. Will the bottle float or sink in water? Density of water is 1 g cm3? Also calculate the mass of the water displaced by the bottle.
Sol. Mass, m = 500 g
Volume, V = 450 cm3
(i) Density of bottle, r = 1.11 g cm3.
Since, density of bottle is greater than the density of water (1 g cm3), so the bottle will sink in water.
(ii) Mass of water displaced by the bottle = Volume of water displaced × Density of water
= Volume of bottle × Density of water
= 450 cm3 × 1 g cm3
= 450 g.

Q14. What is the force of gravitation between two point masses of 1 kg and 2 kg kept 1 m apart?
Sol. m = 1 kg, m2 = 2 kg, r = 1 m
F = 6.67 x 10_11 × = 13.34 × 10_11 N.
This is an extremely small force.

Q15. Calculate the force of gravitation between the earth and the sun.Mass of Earth = 6 × 1024 kg, Mass of Sun = 2 × 1030 kg. The average distance between the two is 1.5 × 1011m.
Sol. F = 35.57 × 1021 N

Q16. Write down the expression for acceleration experienced by a particle on the surface of the moon due to gravitational force on the moon. Find the ratio of this acceleration to that experienced by the same particle on the surface of the earth. If the acceleration due to gravity on the earth is 9.8 ms_2, what is the acceleration of a particle on the moon's surface? Mass of moon = 7.3 × 1022 kg; Mass of Earth = 6 × 1024 kg. Radius of moon= 1.74 × 106 m, Radius of earth = 6.4 × 106 m.
Sol. Acceleration on moon Also acceleration on the surface of Earth
Dividing (1) by (2)

= 0.16
gm =0.16 × gE = 0.16 × 9.8 = 1.57 ms_2

Q17. Find the value of acceleration due to gravity at a height of (a) 6400 km, (b) 12,800 km from the surface of the earth. Radius of earth is 6400 km.
Sol. We know that gh =
Also gh = g
(a) g = 9.8 ms_2, R = 6400 km, R h = 6400 6400 = 12800 km
(gh)A = 9.8 = 9.8 = = 2.45 ms_2
(b) (gh)B = 9.8
= 1.09 ms_2

Q18. A particle is thrown up vertically with a velocity of 50 m/s. (a) What will be its velocity at the highest point of its journey? (b) How high would the particle rise? (c) What time would it take to reach the highest point.
Sol. At the highest point the velocity will be zero.
Considering activity A to B
Using v = u at
0 = 50 _ 9.8 × t
t = 5.1 s
Also v2 _ u2 = 2as
02 _ (50)2 = 2 (_9.8) × s
5 = 127.5 m

Q19. With reference to the above sample problem, (a) Find the time the particle takes from the highest point back to the initial point (b) Find the velocity with which the particle reaches the initial point.
Sol. The data is given in the adjacent figure. Considering activity B to A
Using v2 _ u2 = 2as
v2 _ 02 = 5(9.8)(127.5)
v = 50 m/s
Also v = u at
50 = 0 9.8 (t)
t = 5.1 s

Q20. A ball is dropped from the top of a tower 40 m high. What is its velocity when it has covered 20 m ? What would be its velocity when it hits the ground? Take g= 10 m/s2.
Sol. Let the point B be at a height of 20 m.
Activity from A to B :
u1 = 0, a1 = 10 ms_2, s1 = 20 m, v1 = ?
v12 - u12 = 2a1s1
v12 - 02 = 2 (10) (20)
v12 =202
v1 = 20 m/s
Activity from A to C : C is a point on the ground
u2 = 0, a2 = 10 ms_2, s2 = 40 m, v2 = ?
v22 _ u22 = 2a2s2
v22 _ 02 = 2 (10) (40)
v22 = 800
v2 = 28.28 ms_1

Q21. A body is thrown up with a speed 29.4 ms_1.
(a) What is its speed after (i) t = 1 s, (ii) t = 2 s and (iii) t = 3 s.
(b) What is its height after (i) t = 1 s, (ii) t = 2 s and (iii) t = 3 s.
Sol. (a) (i) u = 29.4 ms_1, a = _9.8 ms_2, t1 = 1 s, v1 = ?
v1 = u at1
= 29.4 (_9.8) x 1 = 19.6 ms_1

(ii) u = 29.4 ms, a = -9.8 ms_2, t2 = 2 s, v2 = ?
v2 = u at2
= 29.4 (_9.8) × 2 = 9.8 ms_1

(iii) u = 29.4 ms_1, a = _9.8 ms_2, t3 = 3 s, v3 = ?
v3 = u at3
= 29.4 (_9.8) × 3 = 0

(b) (i) u = 29.4 ms_1, a = _9.8 ms_2, s1 = h1 , t1 = 1s
h1 = ut1 at12 =29.4 × 1 (_9.8) × 1 =24.5m

(ii) u = 29.4 ms_1, a = -9.8 ms_2, s2 = h2, t2 = 2 s
h2 = ut2 at22 = 29.4 × 2 (_9.8) × 22 = 39.2 m

(iii) u = 29.4 ms_1, a = _9.8 ms_2, s3 = h3, t3 = 3 s
h3 = ut3 at32 = 29.4 x 3 (_9.8) × 32 = 44.1 m

Q22. What is the weight of a person whose mass is 50 kg.
Sol. The weight of the person
W = mg = 50 x 9.8 = 490 N
Note: The gravitational unit of force is kg-f (kilogram force) or kg-wt (kilogram weight) 1 kg-wt = 9.8 N = 1 kg-f
490 N = 50 kg-f

Q23. Weight of a girl is 294 N. Find her mass.
Sol. W = mg
294 = m × 9.8
m = 30 kg

Q24. Weight of an object is 294 N on the surface of the earth. What is its weight at a height of 200 km from the surface of the earth. Radius of the earth= 6400 km.
Sol. Weight at a height of 200 km
Wh = mg
Here mg = 294 N, R = 6400 km, h= 200 km
Wh = 294
= 276.45 N

Note: Weight decreases with increase of height from the surface of the earth.

Q25. The gravitational force between two objects is F. How will this force change when
(i) distance between them is reduced to half ?
(ii) the mass of each object is quadrupled ?

Sol. (i) According to Newton's law of gravitation, gravitational force F between two objects distance r apart is
When distance between them is reduced to half, i.e., r' = r/2, the force,
Thus, F' = 4F
i.e., force becomes 4 times its previous value. "

(ii) Again, according to Newton's law of gravitation, the gravitational force F between two'objects of masses m1 and m2 is
F µ m1m2
When mass of each object is quadrupled,
m'1 = 4m1
and m'2 = 4m2
The force, F ' µ m'1m'2
or F' = 16 F
i.e., force becomes 16 times its previous value.

Q26. A sphere of mass 40 kg is attracted by a second sphere of mass 15 kg when their centres are 20 cm apart, with a force of 0·1 milligram weight. Calculate the value of gravitational constant.
Sol. Here, m1 = 40 kg, m2 = 15 kg
From r = 20cm =2 × 10_1 m
F = 0·1 milligram weight = 0·1 × 10_3 gram weight
= 10_4 × 10_3 kg wt = 10_7 × 9·8 N (1 kg wt = 9.8 N)
From G = 6·53 × 10_11 Nm2/kg2

Q27. Calculate the force of gravity acting on your friend of mass 60 kg. Given mass of earth = 6 x 1024 kg and radius of Earth = 6·4 x 106 m.
Sol. Here, m = 60 kg, M = 6 × 1024 kg
R = 6·4 × 106 m, F = ?
G = 6·67 × 10_11 Nm2/kg2
Thus, F = 58·62 N

Q28. A particle is thrown up vertically with a velocity of 50 m/s. What will be its velocity at the highest point of the journey? How high would the particle rise ? What time would it take to reach the highest point? Take g = 10 m/s2.
Sol. Here, initial velocity, u = 50 m/s
final velocity, v = ?
height covered, h = ?
time taken, t = ?
g = 10 m/s2
At the highest point, final velocity v = 0
From v2 _ u2 = 2 gh, where g = _ 10 m/s2 for upward journey,
0 _ (50)2 = 2 (_10) h
h = 125 m
From v = u gt,
or 0 = 50 (_10) t
t = 50/10 = 5 s

Q29. A force of 15 N is uniformly distributed over an area of 150 m2. Find the pressure in pascals.
Sol. Here, force, F = 15 N
area, A = 150 cm2 = 150 × 10_4 m2 (1 cm = 10_2 m, 1 cm2 = 10_4 m2)
Thus, pressure, P =1000 Pa

Q30. How much force should be applied on an area of 1 cm2 to get a pressure of 15 Pa ?
Sol. Here, area, A = 1 cm2 = 10_4 m2
pressure, P = 15 Pa = 15 N/m2
As P = , F = P × A = (15 N/m2) × (10_4 m2) = 1·5 x 10_3 N

Q31. A block weighing 1·0 kg is in the shape of a cube of length 10 cm. It is kept on a horizontal table. Find the pressure on the portion of the table where the block is kept.
Sol. Here, force acting on the table, F = 1·0 kg = 10 N
area of the table on which this force acts, A = 10 cm × 10 cm = 100 cm2
= 100 × 10_4 m2 = 10_2 m2 (1 cm2 = 10_4 m2)
Pressure on the table, P = 1000 Pa

Exercise -1

Q1. If a rock is brought from the surface of the moon,
(A) its mass will change
(B) its weight will change but not mass

(C) both mass and weight will change
(D) its mass and weight both will remain same

Q2. A body is weighed at the poles and then at the equator. The weight
(A) at the equator will be greater than at the poles
(B) at the poles will be greater than at the equator

(C) at the poles will be equal to the weight at the equator
(D) depends upon the object

Q3. Consider a satellite going round the earth in a circular orbit. Which of the following statements is wrong ?
(A) It is a freely falling body.
(B) It is moving with constant speed.

(C) It is acted upon by a force directed away from the centre of the earth which counter-balances the gravitational pull
(D) Its angular momentum remains constant

Q4. A missile is launched with a velocity less than the escape velocity. The sum of its kinetic and potential energy is
(A) positive
(B) negative
(C) zero

(D) may be positive or negative depending upon its initial velocity

Q5. SI unit of g is
(A) m2/s (B) s/m2 (C) m/s2 (D) m/s

Q6. SI unit of G is
(A) N2_m2/kg (B) N_m2/kg (C) N_m/kg (D) N -m2/kg2

Q7. Choose the correct statement of the following:

(A) All bodies repel each other in this universe.
(B) Our earth does not behave like a magnet.

(C) Acceleration due to gravity is 8.9 m/s2.
(D) All bodies fall at the same rate in vacuum.

Q8. Maximum weight of a body is

(A) at the centre of the earth
(B) inside the earth

(C) on the surface of the earth
(D) above the surface of earth

Q9. If the distance between two masses be doubled, then the force between them will become

(A) times
(B) 4 times
(C) times
(D) 2 times

Q10. A body falls freely towards the earth with
(A) uniform speed
(B) uniform velocity
(C) uniform acceleration
(D) none of these

Q11. If the mass of a body is M on the sufrace of the earth, then its mass on the surface of the moon will be

(A) (B) M
(C) M 6
(D) zero

Q12. Weight
(A) is a vector quantity
(B) of a body in interplanetary space is maximum
(C) increases when the bodies go up
(D) none of these

Q13. The value of g near the earth's surface is
(A) 8.9 m/s2
(B) 8.9 m/s
(C) 9.8 m/s2
(D) 9.8 m/s

Q14. A geostationary satellite

(A) moves faster than the near earth satellite
(B) has a time period less that of a near earth satellite

(C) revolves about the polar axis
(D) is stationary in space

Q15. The force of gravitation between two bodies depend upon
(A) their separation
(B) gravitational constant
(C) product of their masses
(D) all of these

Q16. When an object is thrown up, the force of gravity
(A) acts in the direction of the motion
(B) acts in the opposite direction of the motion

(C) remains constant as the body moves up
(D) increases as the body moves up

Q17. The force of gravitation exists
(A) everywhere in the universe
(B) at the surface of the earth only

(C) inside the earth only
(D) at the surface of the moon only

Q18. 1 kg wt is equal to
(A) 9.8 N
(B) 980N
(C) 98 N
(D) none of these

Q19. 1 kg wt is equal to
(A) 980 dynes
(B) 9.80 dynes
(C) 98 dynes
(D) none of these

Q20. The value of G does not depend on
(A) nature of the interacting bodies
(B) size of the interacting bodies

(C) mass of the interacting bodies
(D) all of these

Match the Column

Q1. Cloumn-I Column-II
(A) Attraction between two planets (p) gravity
(B) Attraction between a body and a planet (q) weightlessness
(C) Free fall  (r) gravitational force
(D) Weight  (s) gravitation

Q2. Cloumn-I Column-II

(A) A body falling freely (p) displacement
(B) Distance with direction (q) velocity
(C) Speed with direction (r) acceleration
(D) Rate of change of velocity (s) uniformly accelerated motion

Q3. Cloumn-I Column-II
(A) Wide straps of school bags (p) motion of planets around the sun.
(B) Archimedes principle (q) weight of an object
(C) Gravitational force  (r) submarine
(D) Upthrust  (s) balloons
(t) to reduce pressure

Q4. Cloumn-I Column-II
(A) Loss in weight  (p) gravitational force
(B) Motion of moon around the earth  (q) upthrust
(C) Airship (r) increasing pressure
(D) Sharp tip of a needle (s) archimedes principle
(t) centripetal force

True & False

1. The force of attraction between two bodies is called gravity.

2. The value of G depends upon the mass of two objects.

3. If a spring balance, holding a heavy object is released, it will read zero weight.

4. The value of G is high if the radius of the body is more and less if radius is less.

5. The centre of mass and centre of gravity for a small body lie at the same point.

6. The gravitational force between two bodies changes if a material body is placed between them.

7. The acceleration of a body thrown up is numerically the same as the acceleration of a downward falling body but opposite is sign.

8. The value of g is zero at the centre of the earth.

9. The inertia of an object depends upon its mass.

10. All objects attract each other along the line joining their centre of mass.

11. Acceleration due to gravity, g = , where symbols have their usual meanings.

12. Relative density has no unit.

13. Archimedes' principle does not apply to gases.

14. Any solid will sink in water if its relative density is less than unity.

Fill in the Blanks

1. .................... is the force of attraction between any two bodies in the universe.

2. .................... is the force of attraction between a body and a planet.

3. Acceleration due to gravity .................... with height from the surface of the earth.

4. Acceleration due to gravity .................... with depth from the surface of the earth.

5. Acceleration due to gravity is a maximum at the ....................

6. .................... of a body changes from place to place but its .................... remains constant.

7. The upward force experienced by a body immersed partially or fully in a fluid is called ....................

8. Density of a substance is defined as the ratio of the mass of a body to its ....................

9. Relative density of a substance is defined as the ratio of the density of the substance to the density of water at ....................

10. In cgs the relative density of a substance is .................... equal to its density in gcm3.

11. The combined weight of the sinker and cork is .................... than the weight of the sinker alone.

12. For a body to float the density of the floating object should be .................... than or equal to the density of the liquid in which it is to float.

13. The centre of ..........is a point, where the total upthrust, due to fluid displaced by part ...........of body acts.

14. A fish .................... by squeezing out air, from its ....................

15. When a body is partly or wholly immersed in a .................... , it experience an ....................

16. An iceberg floats with .................... of its volume below ....................

17. The unit of upthrust in SI is ....................

18. Lactometer is used to measure .................... of ....................

19. A hydrometer sinks .................... in water than in pure milk.

20. The density of hot air is .................... that of cold air.

Answers

Match the Column

1. (A ® s), (B ® p), (C ® q), (D ® r) 2. (A ® s), (B ® p), (C ® q), (D ® r)

3. (A ® t), (B ® r, s), (C ® p, q) (D ® r, s) 4. (A ® q, s), (B ® p,t), (C ® q, s), (D ® r)

True & False

1. False 2. False 3. True 4. True 5. True 6. False 7. True

8. True 9. True 10. True 11. True 12. True 13. False 14. False

Fill in the Blanks

1. gravitation 2. gravity 3. decreases 4. decreases

5. poles 6. weight, mass 7. buoyant force 8. volume

9. 4°C 10. numerically 11. less 12. less

13. buoyancy, the immersed 14. dives, floating tube 15. fluid, upthrust 16. 11/12, water

17. Newton 18. purity, milk 19. more 20. less than

Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

## Science Class 9

90 videos|225 docs|135 tests

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;