NCERT Exemplar Solutions for Class 11 Physics Chapter 8 Gravitation is an essential learning material for the students of CBSE Class 11 and for those who are preparing for Engineering and Medical entrance examinations. These NCERT Exemplar Solutions Class 11 Physics Chapter 8 PDF contain questions from NCERT exemplar books in addition to important derivations and solved numerical questions, which consist of MCQs, worksheets, HOTS and short and long answer questions.
Gravitation is a phenomenon that occurs in nature, where objects, such as planets, stars and even galaxies with energy or mass, are attracted to each other. This force of attraction that exists between two objects causes them to gravitate towards each other. Albert Einstein initially proposed the general theory of relativity, describing gravity most accurately. Students can download and practise NCERT Exemplar Solutions for Class 11 Physics Chapter 8 Gravitation from the link below to score high in the annual exam.
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Multiple-choice Questions I
8.1. The earth is an approximate sphere. If the interior contained matter which is not of the same density everywhere, then on the surface of the earth, the acceleration due to gravity
a) will be directed towards the centre but not the same everywhere
b) will have the same value everywhere but not directed towards the centre
c) will be the same everywhere in magnitude directed towards the centre
d) cannot be zero at any point
Answer:
The correct answer is d) cannot be zero at any point
8.2. As observed from earth, the sun appears to move in an approximately circular orbit. For the motion of another planet like mercury as observed from earth, this would
a) be similarly true
b) not be true because the force between earth and mercury is not inverse square law
c) not be true because the major gravitational force on mercury is due to the sun
d) not be true because mercury is influenced by forces other than gravitational forces
Answer:
The correct answer is c) not be true because the major gravitational force on mercury is due to the sun
8.3. Different points in the earth are at slightly different distances from the sun and hence experience different forces due to gravitation. For a rigid body, we know that if various forces act at various points in it, the resultant motion is as if a net force acts on the cm, causing translation and a net torque at the cm causing translation and a net torque at the cm causing rotation around an axis through the cm. For the earth-sun system
a) the torque is zero
b) the torque causes the earth to spin
c) the rigid body result is not applicable since the earth is not even approximately a rigid body
d) the torque causes the earth to move around the sun
Answer:
The correct answer is a) the torque is zero
8.4. Satellites orbiting the earth have a finite life, and sometimes debris of satellites fall to the earth. This is because
a) the solar cells and batteries in satellites run out
b) the laws of gravitation predict a trajectory spiralling inwards
c) of viscous forces causing the speed of the satellite and hence height to gradually decrease
d) of collisions with other satellites
Answer:
The correct answer is c) of viscous forces causing the speed of the satellite and hence height to gradually decrease
8.5. Both the earth and moon are subject to the gravitational force of the sun. as observed from the sun, the orbit of the moon
a) will be elliptical
b) will not be strictly elliptical because the total gravitational force on it is not central
c) is not elliptical but will necessarily be a closed curve
d) deviates considerably from being elliptical due to the influence of planets other than earth
Answer:
The correct answer is b) will not be strictly elliptical because the total gravitational force on it is not central
8.6. In our solar system, the interplanetary region has chunks of matter called asteroids. They
a) will not move around the sun since they have very small masses compared to the sun
b) will move in an irregular way because of their small masses and will drift away from outer space
c) will move around the sun in closed orbits but not obey Kepler’s laws
d) will move in orbits like planets and obey Kepler’s laws
Answer:
The correct answer is d) will move in orbits like planets and obey Kepler’s laws
8.7. Choose the wrong option.
a) inertial mass is a measure of the difficulty of accelerating a body by an external force, whereas gravitational mass is relevant in determining the gravitational force on it by an external mass
b) that the gravitational mass and inertial mass are equal is an experimental result
c) that the acceleration due to gravity on earth is the same for all bodies is due to the equality of gravitational mass and inertial mass
d) gravitational mass of a particle-like proton can depend on the presence of neighbouring heavy objects, but the inertial mass cannot
Answer:
The correct answer is d) gravitational mass of a particle-like proton can depend on the presence of neighbouring heavy objects, but the inertial mass cannot
8.8. Particles of masses 2M, m and M are respectively at points A, B, and C with AB = 1/2 (BC). M is much smaller than M, and at time t = 0, they are all at rest. At subsequent times before any collision takes place
a) m will remain at rest
b) m will move towards M
c) m will move towards 2M
d) m will have oscillatory motion
Answer:
The correct answer is c) m will move towards 2M
Multiple Choice Questions II
8.9. Which of the following options is correct?
a) acceleration due to gravity decreases with increasing altitude
b) acceleration due to gravity increases with increasing depth
c) acceleration due to gravity increases with increasing latitude
d) acceleration due to gravity is independent of the mass of the earth
Answer:
The correct answers are
a) acceleration due to gravity decreases with increasing altitude
c) acceleration due to gravity increases with increasing latitude
8.10. If the law of gravitation, instead of being inverse-square law, becomes an inverse-cube-law
a) planets will not have elliptic orbits
b) circular orbits of planets are not possible
c) projectile motion of a stone thrown by hand on the surface of the earth will be approximately parabolic
d) there will be no gravitational force inside a spherical shell of uniform density
Answer:
The correct answers are
a) planets will not have elliptic orbits
b) circular orbits of planets are not possible
c) projectile motion of a stone thrown by hand on the surface of the earth will be approximately parabolic
8.11. If the mass of the sun were ten times smaller and gravitational constant G was ten times larger in magnitude
a) walking on the ground would become more difficult
b) the acceleration due to gravity on earth will not change
c) raindrops will fall much faster
d) aeroplanes will have to travel much faster
Answer:
The correct answers are
a) walking on the ground would become more difficult
c) raindrops will fall much faster
d) aeroplanes will have to travel much faster
8.12. If the sun and the planets carried huge amounts of opposite charges,
a) all three of Kepler’s laws would still be valid
b) only the third law will be valid
c) the second law will not change
d) the first law will still be valid
Answer:
The correct answers are
a) all three of Kepler’s laws would still be valid
c) the second law will not change
d) the first law will still be valid
8.13. There have been suggestions that the value of the gravitational constant G becomes smaller when considered over a very large time period in the future. If that happens to our earth,
a) nothing will change
b) we will become hotter after billions of years
c) we will be going around but not strictly in closed orbits
d) after a sufficiently long time, we will leave the solar system
Answer:
The correct answers are
c) we will be going around but not strictly in closed orbits
d) after a sufficiently long time, we will leave the solar system
8.14. Supposing Newton’s law of gravitation for gravitation forces F1 and F2 between two masses m1 and m2 at positions r1 and r2 read
where Mo is a constant of the dimension of mass, r12 = r1 – r2 and n is a number. In such a case,
a) the acceleration due to gravity on earth will be different for different object
b) none of the three laws of Kepler will be valid
c) only the third law will become invalid
d) for n negative, an object lighter than water will sink into the water
Answer:
The correct answers are
a) the acceleration due to gravity on earth will be different for different object
c) only the third law will become invalid
d) for n negative, an object lighter than water will sink into the water
8.15. Which of the following is true?
a) a polar satellite goes around the earth’s pole in a north-south direction
b) a geostationary satellite goes around the earth in an east-west direction
c) a geostationary satellite goes around the earth in a west-east direction
d) a polar satellite goes around the earth in an east-west direction
Answer:
The correct answers are
a) a polar satellite goes around the earth’s pole in a north-south direction
c) a geostationary satellite goes around the earth in a west-east direction
8.16. The centre of mass of an extended body on the surface of the earth and its centre of gravity
a) are always at the same point for any size of the body
b) are always at the same point only for spherical bodies
c) can never be at the same point
d) is close to each other for objects, say of sizes less than 100 m
e) both can change if the object is taken deep inside the earth
Answer:
The correct answer is d) is close to each other for objects, say of sizes less than 100 m
Very Short Answers
8.17. Molecules in the air in the atmosphere are attracted by the gravitational force of the earth. Explain why all of them do not fall into the earth, just like an apple falling from a tree.
Answer:
Molecules in the air in the atmosphere are attracted by the gravitational force of the earth, but they do not fall into the earth because they are in random motion, whereas apple has a downward motion.
8.18. Give one example of each of the central force and non-central forces.
Answer:
Central force: Electrostatic force acting on the point charge
Non-central force: Nuclear force between the atoms
8.19. Draw areal velocity versus time graph for mars.
Answer:
According to Kepler’s law, the areal velocity of mars around the sun does not change when it revolves around the sun. therefore, the graph of areal velocity versus time is a straight line.
8.20. What is the direction of the areal velocity of the earth around the sun?
Answer:
The direction of the areal velocity of the earth around the sun is in the direction of the product of r and v.
8.21. How is the gravitational force between two point masses affected when they are dipped in the water keeping the separation between them the same?
Answer:
The gravitational force is unaffected by the medium, and it remains the same between the point masses irrespective of their surrounding. Therefore, when the two-point masses are dipped in water, the gravitational force between them remains the same.
8.22. Is it possible for a body to have inertia but no weight?
Answer:
Yes, it is possible for a body to have inertia, but no weight, as inertia is associated with the mass of the body. A satellite revolving around the earth is an example of a body with inertia and has no weight.
8.23. We can shield a charge from electric fields by putting it side a hollow conductor. Can we shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means?
Answer:
No, it is possible to shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere as gravitation is independent of the medium.
8.24. An astronaut inside a small spaceship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity?
Answer:
When the size of the space station orbiting around the earth is large, the astronaut will experience variation and this is because of acceleration due to gravity.
8.25. The gravitational force between a hollow spherical shell and a point mass is F. Show the nature of F vs r graph where r is the distance of the point from the centre of the hollow spherical shell of uniform density.
Answer:
R is the radius of the spherical shell and r is the distance between m and M. M is the mass of the hollow spherical shell and m is the mass of the point. We know that,
F = GMm/r2
When the F = 0, F = GM/r2 for r ≥ R
Below is the graph which shows the variation of F versus r such that when force is maximum, r tends to infinity.
8.26. Out of aphelion and perihelion, where is the speed of the earth more and why?
Answer:
According to Kepler’s second law, a real velocity is constant and is given as:
rp × vp = rA × vA
rA/rp = vp/vA
rA > rp and vp > vA
8.27. What is the angle between the equatorial plane and the orbital plane of
a) polar satellite?
b) geostationary satellite?
Answer:
a) The angle between the equatorial plane and the orbital plane of the polar satellite is 90o
b) The angle between the equatorial plane and the orbital plane of a geostationary satellite is 0o
Short Answer
8.28. Mean solar day is the time interval between two successive noon when the sun passes through the zenith point. The sidereal day is the time interval between two successive transits of a distant star through the zenith point. Drawing the appropriate diagram showing earth’s spin and orbital motion, show that the mean solar day is four minutes longer than the sidereal day. In other words, distant stars would rise 4 minutes early every successive day.
Answer:
E and E’ is the polar axis of the earth and its movement.
P’ is the translational motion
In every 24 hours, the orbit of the earth is approximately advanced by 1o
Therefore, time taken for transverse is = (24/360o) (1o) = 239.3 sec = 4 minutes.
8.29. Two identical heavy spheres are separated by a distance 10 times their radius. Will an object placed at the midpoint of the line joining their centres be in a stable equilibrium or unstable equilibrium? Give a reason for your answer.
Answer:
M is the mass of the spheres and R is the radius of the spheres.
P is the midpoint of A and B.
The magnitude of the force is given as:
F1 = F2 = GMm/(5R)2
The equilibrium will be unstable as the resultant force acting on the object moves towards sphere A.
8.30. Show the nature of the following graph for a satellite orbiting the earth.
a) KE vs orbital radius R
b) PE vs orbital radius R
c) TE vs orbital radius R
Answer:
a) K = 1/2mv2 = (1/2 m)(GM/R)
b) PE of satellite is U = -GMm/R = -2K
c) Total energy of the satellite = -GMm/2R
8.31. Shown are several curves. Explain with reason, which ones amongst them can be possible trajectories traced by a projectile
Answer:
The given options, option (c) show the focus of trajectory.
8.32. An object of mass m is raised from the surface of the earth to a height equal to the radius of the earth, that is, taken from a distance R to 2R from the centre of the earth. What is the gain in its potential energy?
Answer:
The PE of the body on the earth = -GMm/R
When the body is close to the equator, the PE = -GMm/2R
Gain in PE = 1/2 mgR
8.33. A mass m is placed at P a distance h along the normal through the centre O of a thin circular ring of mass M and radius r. If the mass is removed further away such that OP becomes 2h, by what factor the force of gravitation will decrease, if h = r?
Answer:
Let r be the radius of the ring and m be the mass of the ring. Small element dM is considered as the mass, the gravitation force dF = G(dM)m/x2
Where x2 = r2 + h2
When the force is integrated, the distance between the m and the ring is 2h.
Long Answers
8.34. A star, like the sun, has several bodies moving around it at different distances. Consider that all of them are moving in circular orbits. Let r be the distance of the body from the centre of the star and let its linear velocity be v, angular velocity ω, kinetic energy K, gravitational potential energy U, total energy E, and angular momentum l. As the radius r of the orbit increases, determine which of the above quantities increase and which ones decrease.
Answer:
8.35. Six-point masses of mass m each are at the vertices of a regular hexagon of side l. Calculate the force on any of the masses.
Answer:
AE = AG + EG = AG + AG = 2AG = 2l cos 30o
AE = AC = √3l
AD = 2l
Therefore,
Force on A due to B is Gm2/l2 along B to A
Force on A due to C is Gm2/3l2 along D to A
Force on A due to D is Gm2/4l2 along D to A
Force on A due to E is Gm2/3l2 along E to A
Therefore, total force = F = Gm2/l2 (1 + 1/√3 + 1/4) along DA
8.36. A satellite is to be placed in equatorial geostationary orbit around the earth for communication
a) calculate height of such a satellite
b) find out the minimum number of satellites that are needed to cover the entire earth so that at least one satellite is visible from any point on the equator
Answer:
a) Mass of the earth, M = 6 1024 kg
Radius of the earth, R = 6.4 103 m
Time period = 24 36 102 s
G = 6.67 10-11 N.m2/kg2
Orbital radius = R + h
Using orbital velocity, we can calculate h = 35,940 km
b) The no.of satellites that are needed to cover the entire earth so that at least one satellite is visible from any point of the equator is 3.
8.37. Earth’s orbit is an ellipse with an eccentricity of 0.0167. Thus, the earth’s distance from the sun and speed as it moves around the sun varies from day to day. This means that the length of the solar day is not constant throughout the year. Assume that the earth’s spin axis is normal to its orbital plane and find out the length of the shortest and the longest day. A day should be taken from noon to noon. Does this explain the variation in the length of the day during the year?
Answer:
Velocity of the earth at perigee = vp
Velocity of the earth at apogee = va
Angular velocity of the earth at perihelion = êž·p
Angular velocity of the earth at aphelion = êž·a
êž·p/êž·a = 1.0691
êž· is the mean angular speed = 1.0691
êž·p/êž· = êž·/êž·a = 1.034
Therefore, it can be said that this does not explain the actual variation in the length of the day during the year.
8.38. A satellite is in an elliptic orbit around the earth with an aphelion of 6R and perihelion of 2R where R = 6400 km is the radius of the earth. Find eccentrically of the orbit. Find the velocity of the satellite at apogee and perigee. What should be done if this satellite has to be transferred to a circular orbit of radius 6R?
Answer:
Radius of perigee = rp = 2R
Radius of apogee = ra = 6R
rp = 2R
ra = 6R
rp = a(1-e) = 2R
ra = a(1+e) = 6R
From the above equation, e = 1/2
Since there is no external force on the system, from the law of conservation of angular momentum
L1 = L2
ma = mp = m = mass of the satellite
va/vp = rp/ra = 1/3
Therefore, vp = 3va
From the conservation of energy, we get
va = 2.28 km/s
vc = 3.23 km/s
The velocity at the apogee is vc – va = 0.95 km/s
Important Topics Covered in Chapter 8 Gravitation
- Kepler’s laws
- The universal law of gravitation
- The gravitational constant
- Acceleration due to the gravity of the earth
- Acceleration due to gravity below and above the surface of the earth
- Gravitational potential energy
- Escape speed
- Earth satellites
- The energy of an orbiting satellite
- Geostationary and polar satellites
- Weightlessness.
| Also Access |
| NCERT Solutions for Class 11 Physics Chapter 8 |
| CBSE Notes for Class 11 Physics Chapter 8 |
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