Physics numerical of chapter 10 gravitation class 9

Physics numerical of chapter 10 gravitation class 9- Part-01

1. 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?

2. A sphere of mass 40kg is attracted by a second sphere of mass 15kg when their centers are 20 cm apart, with a force of 0.1 milligram weight. Calculate the value of gravitational constant.

3. A body of mass 1 kg is placed at a distance of 2m from another body of mass 10kg. At what distance from the body of 1 kg, another body of mass 5 kg be placed so that the net force of gravitation acting on the body of mass 1 kg is zero?

4. A geostationary satellite is orbiting the earth at a height 5 R above the surface of earth, where R is the radius of earth. Find the time period of another satellite at a height of 2 R from the surface of earth.

5. The distance of planet Jupiter from the sun 5.2 times that of Earth. Find the period of revolution of Jupiter around sun.

6. If the distance of Earth from the Sun were half the present value, how many days will make one year?

7. Two satellites of a planet have periods 32 days and 256 days. If the radius of orbit of former is R, find the orbital radius of the latter.

8. The mass of Earth is 6 x 10²⁴ kg and that of moon is 7.4 x 10²² kg. If the distance between the Earth and the Moon is 3.84 x 105 km, calculate the force exerted by Earth on the Moon. Given G = 6.7 x 10^–11 Nm2/kg².

9. If the distance between two masses is increased by a factor of 4, by what factor would the mass of one of them have to be altered to maintain the same gravitational force?

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

11. The mass of Sun is 2 x 1030 kg and mass of Earth is 6 x 1024 kg. If the distance between the centres of Sun and Earth is 1.5 x 108 km, calculate the force of gravitation between them.

12. Two electrons each of mass 9.1 x 10–31 kg are at a distance of  10A⁰. Calculate the gravitational force of attraction between them. Given 1A⁰ =10^-10m

13. The gravitational force between force two objects is 100 N. How should the distance between these objects be changed so that force between them becomes 50 N?

14. Calculate the force of gravitation between two objects of masses 80kg and 1200 kg kept at a distance of 10 m from each other. Given G = 6.67 x 10–11 Nm2/kg².

15. Calculate the force of attraction between the Earth and the Sun, given that the mass of Earth is 6 x 1024 kg and that of sun is 2 x 1030 kg. The average distance between the two is 1.5 x 10¹¹m.

Physics numerical of chapter 10 gravitation class 9- Part-02

1. Calculate the force of gravity acting on your friend of mass 60kg. Given mass of earth = 6 x 1024 kg and radius of Earth = 6.4 x 106m.

2. Mass of an object is 10kg. What is its weight on Earth?

3. What is the mass of an object whose weight is 49N?

4. An object weighs 10N when measured on the surface of the earth. What would be its weight when measured on the surface of the Moon?

5. An object is thrown vertically upwards and rises to a height of 10m. Calculate (i) the velocity with which the object was thrown upwards and (ii) the time taken by the object to reach the highest point.

6. A force of 2 kg wt. acts on a body of mass 4.9kg. Calculate its acceleration.

7. A force of 20N acts upon a body weight is 9.8N. What is the mass of the body and how much is its acceleration?

8. A body has a weight of 10 kg on the surface of earth. What will be its mass and weight when taken to the centre  of earth?

9. How much would a 70 kg man weigh on moon? What will be his mass on earth and moon? Given g on moon = 1.7 m/s2.

10. The Earth’s gravitational force causes an acceleration of 5 m/s2 in a 1 kg mass somewhere in space. How much will the acceleration of a 3 kg mass be at the same place?

11. A particle is thrown up vertically with a velocity of 50m/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.

12. If a planet existed whose mass was twice that of Earth and whose radius 3 times greater, how much will a 1kg mass weigh on the planet?

13. A boy on cliff 49m high drops a stone. One second later, he throws a second stone after the first. They both hit the ground at the same time. With what speed did he throw the second stone?

14. A stone drops from the edge of a roof. It passes a window 2m high in 0.1s. How far is the roof above the top of the window?

15. A stone is dropped from the edge of a roof. (a) How long does it take to fall 4.9m ?

(b) How fast does it move at the end of that fall? (c) How fast does it move at the end of 7.9m? (d) What is its acceleration after 1s and after 2s?

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Solved Question on Class IX » Science » Gravitation » The Universal Law Of Gravitation.

Q.1. When we move from the poles to the equator. Hence, the value of ‘ g ’ decreases.  Why

Ans: The shape of earth is an ellipse so when we move from the poles to the equator the radius of the earth R increases. Hence, the value of ‘ g ’ decreases because value 'g' is inversely proportional to the radius of earth. g = GM/R2

Q. 2. What is the diffrence between centrifugal force and centripetal force?

Ans: Centripetal Force

(i) It is the force that keeps a body in circular path. (ii) It acts toward the center.

Centrifugal Force

(i) It is the pseudo force that tries to make a body fly off the circular path.  (ii) It acts outward the center.

Q.3. Explain :Centrifugal force and Centripetal force?

Ans: A force which is required to move a body uniformly in a circle is known as centripetal force. This force acts along the radius and towards the center of the circle,

Centrifugal force arises when a body is moving actually along a circular path, by virtue of tendency of the body to regain its natural straight line path. This force acts along the radius and away from the center of the circle.

Q.4 an astronaut has 80 kg mass on earth  (a)what is his weight on earth? (b) what will be his mass and weight on mars where g=3.7 m/s2

Ans: Mass of astronaut = 80 kg

Weight on earth = mg = (80)(9.8) N = 784 N  Weight on mars = mg' = (80)(3.7) N = 296 N

Q.5. A certain particle has a weight of 30N at a place where the acceleration due to gravity is 9.8 m/s2

(a) What are its mass and weight at a place where acceleration due to gravity is 3.5 m/s square

(b) What are its mass and weight at a place where acceleration due to gravity is 0

Ans (a) Weight of the body, W = 30 N =mg   Mass of the body, m = W/g =30 /9.8= 3.06 kg

New weight of the body, W' = mg' = (3.06) (3.5) N = 10.71 N

(b) . Mass remains the same but weight becomes zero.

Q.6. Derive the inverse square of Newton.

Ans: Let a planet of mass m revolves around the sun of mass M in nearly circular orbit of radius r, with a constant angular velocity ω. Let T be the time period of the revolution of the planet around the sun. then

w = 2p/T

The centripetal force acting on the planet, F = mrw2 = mr (2p/T)2 = (4p2mr)/T2  ------(i)

According to Keller’s third law

“The square of the time period of revolution of a planet around the sun is directly proportional to the cube of semi major axis of its elliptical orbit”

T2    µ r3

T2    = K r3  --------------(ii)

Here, K is proportionality constant.

from (i)and (ii)

F = (4p2mr)/ K r3

F = (4p2/ K)x {m/ r2 }

F µ m/r2

According to Newton, the gravitational attraction between the sun and the planet is mutual. So if F depends upon the mass of the planet m then it should also be directly proportional to the mass of the sun, M.

Hence,  4p2/ K µ M

4p2/ K = G M

F = G (Mm/r2)

This is Newton’s law of gravitation.

Q,7. What is the difference between gravity and gravitation?

Ans: Gravity is defined as the ability of earth to attract another body by virtue of their masses. Gravitation is the phenomenon which explains the force of attraction between two masses separated by a certain distance. This force is known as Gravitational Force

Q.8.What are these  :(( i) Product Rule (ii) Inverse Square rule (iii) Universal gravitational constant iv) Universal law of gravitation:

Ans: (i) Product rule: Force between two mass separated by a distance is directly proportional to the product of the two masses.

(ii) Inverse square law means that the force is inversely proportional to the square of the distance between two objects. Gravitational force is an example of inverse square law. The relation between the force of gravitation and distance is F∝1/r2

(iii)Universal gravitational constant: The constant of proportionality is called the universal gravitational constant. Gravitational constant is defined as the force of attraction between two unit masses kept at unit distance.  For example if we choose m 1, m 2 such that, m 1 = m 2 = 1 and keep them at a unit distance (r =1), gravitational constant is equal to gravitation force of attraction between them

(iv) Universal law of gravitation: a force of attraction between two masses separated by some distance. The gravitational force between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

IX Term -1 Solved numerical Gravitation and Laws of motion

Q.A rifle of mass 3 kg fires a bullet of mass 0.03 kg. The bullet leaves the barrel of the rifle at a velocity of 100 m/s. If the bullet takes 0.003 s to move through its barrel, calculate the force experienced by the rifle due to the recoil.

Ans:    Mass of the rifle (m₁₎ = 3 kg  Mass of the bullet (m₂₎ = 0.03 kg
Initial velocity of the rifle (u₁) = u₂ = 0,   Final velocity of the rifle (v₁) = ?
Final velocity of the bullet (v₂) = 100 m s^-1
According to law of conservation of momentum = m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Using the law of conservation of momentum,  0 + 0 = 3v₁ + 0.03 x 100 =-1m/s 
Negative sign shows that the rifle moves in a direction opposite to that of the bullet,
Force experienced by the rifle due to its recoil= m1v1/t =(3x-1)/0.003=-1000N 
Therefore, the person would experience a force of 1000 N in the backward direction due to the recoil of the rifle.

 Q. Action and reaction forces is never equal to zero. Why?

Ans: The action and reaction force are equal and opposite but their resultant is never equal to zero or cancel because The action and reaction force are acting on two different bodies.

Q. A bullet leaves a rifle with a velocity of 100m/s and the rifle of mass 2.5 kg recoils with a velocity of 1m/s. Find the mass of the bullet?

Ans: According to law of conservation of momentum = m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Þ 0=2.5x1 +m2 x100  Þ 2.5/100= the mass of the bullet  Þ 25gm= the mass of the bullet

Q. A body of mass 5 kg undergoes a change in speed from 30 to 40m/s . Calculate its increase in momentum?

Ans: Increase in momentum =m(v-u) =5 kg(40-30)m/s=50kgm/s

Q. Two bodies of mass 10 kg and 12 kg are falling freely. What is the acceleration produced in the bodies due to force of gravity?

Ans: 9.8 m/s².ast he acceleration due to gravity produced in both the bodies is the same as it is independent of the mass of the body.

Q. What will happen to the force of gravitation between two objects A and B if the distance between them is reduced to half?

Ans: F=1/r²

F’ = 1/(1/2r)² =4(1/r²) =4(F)

if the distance between two objects A and B is reduced to half the force of gravitation between two objects A and B increases 4 times

Q. What will happen to the force of gravitation between two objects A and B If the mass of the object A is doubled.

Ans: F = m M

F’= 2mxM=2(mM)=2F

the force of gravitation between two objects A and B increases 2 times

Q. Show that universal gravitational constant is nothing but force of gravitation between two unit masses separated by unit distance.

Ans: F=GmM/r²

if m=M=1kg and r =1m

then, F = G i.e., gravitational constant is equal to the force of gravitation.

Q. A boy drops a stone from a cliff, which reaches the ground in 20 seconds. Calculate the height of the cliff.

Ans: the height of the cliff.=S=ut+1/2at² =0+1/2x10x20x20=2000m=2km

CBSE (9th)IX Physics Test Paper Ch:03 Gravitation

Class:   IX Subject:  Physics  ASSIGNMENT Ch:03 GRAVITATION

1. Define: gravitation, gravity and gravitational force.

2. State the universal law of gravitation and its mathematical form.

3. In what source is ‘G’ universal?

4. What happens to the force of attraction between two objects when

(i) Their mass are halved? (ii) Distance between them is increased to 4 times its previous value. (iii) Distance between them as well as each of the mass is increase to 4 times.

5. If the distance between two bodies is increased 4 times by what factor should the mass of the bodies be altered so that the gravitational force between them remains the same?

6. What is the force between two spheres weighing 20 kg each and placed 50 cm apart?

7. A sphere of mass 40 kg is attracted by another sphere of mass 15 kg when their centers are  0.2 m  apart with  force of     9.8 x 10-7

N. Calculate value of ‘G’.

8. Find the distance between two stones each of mass 2 kg so that the gravitational force between them is 1N.

9. Gravitational force between two objects on earth is 2N. What will be the gravitational force between these  two objects on the surface of the earth?

10. Why don’t we see objects in the universe colliding or moving towards each other due to gravitational force?

11. Why do all objects fall towards the earth?

12. Give a few examples / applications of the universal law of gravitation.

13. On what factor [s} does the gravity of a planet depend?

14. Define ‘G’ and give its value.

15. Differentiate between ‘g’ and ‘G’.

16. Define ‘g’ and give its value on the surface of the earth.

17. Name the factors on which ‘g’ depends.

18. Calculate the gravitational force between a body of mass 100 kg and the earth. Also calculate the acceleration produced in the body and that in earth.

19. A body weighs 1 kg on the surface of the moon. If mass of the moon is 7.4 x 10 22 kg and radius of moon is 1740 km. Calculate:

a. The force acting between the body & the moon.

b. Acceleration produced in the body

c. Acceleration produced in moon.

20. A planet has mass and radius 1/3 of those of earth. Calculate the acceleration due to gravity of the planet and compare it with acceleration due to gravity on earth. If an object has 5 kg mass on earth. Calculate its weight on the planet.

21. Differentiate between mass & weight of a body.

22. A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate:

(I). the max height to which it rises.  (ii) Total time it takes to return to surface of the earth.

23. A stone is released from the top of a tower of height 19.6 m. Calculate the final velocity of a body just before touching the ground.

24. A stone is thrown vertically upwards with an initial velocity of 40m/s. Find the max height reached by the  stone. What is the net displacement and the total distance covered by the stone? [g = 10m/s2]

25. The object dropped from a height (h) with initial velocity zero strikes the ground with a velocity of 30m/s.  How long does it take to reach the ground. Also find h [g = 10m/s2]

26. The mass of a man is 75 kg. What will be his weight on surface of earth? What will be his weight on surface of moon?

27. A ball is projected vertically upwards with a certain velocity. What is its acceleration?

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