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GRAVITATION:--
OBJECTIVE - I
1. The acceleration of the moon with respect to earth is 0.0027 m/s² and the acceleration of an apple falling on the earth's surface is about 10 m/s². Assume that the radius of the moon is one-fourth of the earth's radius. If the moon is stopped for an instant and then released, it will fall towards the earth. The initial acceleration of the moon towards the earth will be
(a) 10 m/s² (b) 0.0027 m/s² (c) 6.4 m/s² (d) 5.0 m/s²
ANSWER: (b)
EXPLANATION: When moving in the orbit, the moon balances the force on it by the earth with its uniform circular motion in the orbit resulting in centrifugal force. When stopped for an instant, the force on the moon by the earth does not change, only balancing force is absent. So acceleration at that instant = Force /mass remains the same. Only in the orbit, it is the instantaneous centripetal acceleration.
2. The acceleration of the moon just before it strikes the earth in the previous question is
(a) 10 m/s² (b) 0.0027 m/s² (c) 6.4 m/s² (d) 5.0 m/s²
ANSWER: (c)
EXPLANATION: When the moon strikes the earth its center is R/4 above the surface of the earth. So the acceleration of the moon will be calculated at R+R/4 = 5R/4 away from the center of the earth. Since the acceleration is inversely proportional to the square of the distance, the acceleration of the moon at the time of strike
= g*{R/(5R/4)}² = g*(4/5)² = 16g/25 =160/25 =6.4 m/s².
3. Suppose the acceleration due to gravity at the earth's surface is 10 m/s² and at the surface of Mars it is 4.0 m/s². A 60 kg passenger goes from the earth to the Mars in a spaceship moving with a constant velocity. Neglect all other objects in the sky. Which part of the figure (11-Q1) best represents the weight (net gravitational force) of the passenger as a function of time.
(a) A (b) B (c) C (d) D
ANSWER: (c)
EXPLANATION: Since the acceleration due to gravity varies inversely to the square of the distance hence the apparent weight (net gravitational force) of the passenger with respect to time will not be a straight line but a curve. In between the earth and the mars, there will be a point where the gravitational force due to both of the bodies will be equal and opposite, hence the apparent weight will zero but never negative. Out of the three curves in the figure, only curve C fulfills this condition.
4. Consider a planet in some solar system which has a mass double the mass of the earth and density equal to the average density of the earth. An object weighing W on the earth will weigh
(a) W (b) 2W (c) W/2 (d) 2¹/³ W at the planet.
ANSWER: (d)
EXPLANATION: Since the density is same the volume and hence the radius of the planet will be more than the radius of the earth. Let the radius of the earth be R and that of the planet be R'. If the density is ρ, then
4πR'³ρ/3 = 2*4πR³ρ/3
→R'³ =2R³
→R' = 2¹/³R
Acceleration due to gravity on this planet = G*2M/R'²
(M is mass of the earth)
=2GM/(2¹/³R)²
=2⁽¹⁻²/³⁾(GM/R²)
=2¹/³g
Hence the weight on the planet =2¹/³*mg =2¹/³W.
5. If the acceleration due to gravity at the surface of the earth is g, the work-done in slowly lifting a body of mass m from the earth's surface to a height R equal to the radius of the earth is
(a) ½mgR (b) 2mgR (c) mgR (d) ¼mgR
ANSWER: (a)
EXPLANATION: The distance of the body from the center of the earth at a height R from the earth's surface = 2R.
The acceleration due to gravity at distance x from the center of the earth ( for x>R) = GM/x²
Force on the body by the earth =mGM/x²
Since the body is lifted slowly, hence the moving force is ≈ mGM/x²
Hence the work done by the force in taking the body from R to 2R
= ∫m(GM/x²)dx
(limit of integration from R to 2R)
=mGM[-1/x]
=mGM[-1/2R+1/R]
=mGM/2R =½m(GM/R²)R = ½mgR
6. A person brings a mass of 1 kg from infinity to a point A. Initially, the mass was at rest but it moves at a speed of 2 m/s as it reaches A. The work done by the person on the mass is -3 J. The potential at A is
(a) -3 J/kg (b) -2 J/kg (c) -5 J/kg (d) None of these
ANSWER: (c)
EXPLANATION: Work done by the person = -3 J (Given)
Work done in increasing the velocity from rest to 2 km/s = K.E. of the mass at A =½mv² = ½*1*2² = 2 J
Let the potential at A be X J/kg,
So, the change in P.E. of the mass = the work done in slowly bringing the mass from infinity to A = mX
= 1*X = X J
So the total work done = Change in P.E + Change in K.E. = X+2 J, but it is given -3 J
so, X+2 = -3
→X = -3-2 = -5 J
7. Let V and E be the gravitational potential and the gravitational field at a distance r from the center of a uniform shell. Consider the following two statements:
(A) the plot of V against r is discontinuous.
(B) the plot of E against r is discontinuous.
(a) Both A and B are correct.
(b) A is correct but B is wrong.
(c) B is correct but A is wrong.
(d) Both A and B are wrong.
ANSWER: (c)
EXPLANATION: (A) is not correct because the gravitational potential inside the shell is constant =-GM/a, where a is the radius of the shell. As the distance increases from a, the potential gradually starts increasing till infinity to a value zero. So the curve is not discontinuous.
(B) is correct because the gravitational field inside the shell is zero and just as it goes outside the value jumps to GM/a² and then gradually decreases to zero at infinity. So the curve at r =a is discontinuous.
8. Let V and E be the gravitational potential and the gravitational field at a distance r from the center of a uniform solid sphere. Consider the following two statements:
(A) the plot of V against r is discontinuous.
(B) the plot of E against r is discontinuous.
(a) Both A and B are correct.
(b) A is correct but B is wrong.
(c) B is correct but A is wrong.
(d) Both A and B are wrong.
ANSWER: (d)
EXPLANATION: (A) is wrong because the value of V at the center is -3GM/2a and it increases continuously to -GM/a at the surface and from -GM/a at the surface to zero at the infinity. So it is continuous.
(B) is wrong because the value of E increases linearly from zero at the center to GM/a² at the surface, then decreases from GM/a² at the surface to zero at the infinity. So it is also continuous.
9. Take the effect of bulging of the earth and its rotation in the account. Consider the following statements:
(A) There are points outside the earth where the value of g is equal to its value at the equator.
(B) There are points outside the earth where the value of g is equal to its value at the poles.
(c) B is correct but A is wrong.
(d) Both A and B are wrong.
ANSWER: (b)
EXPLANATION: As we move towards poles the value of g starts increasing. Another fact is that as we go up at a place the value of g starts decreasing. So at any point on the surface of the earth where the value of g is more than at the equator, there will be points outside the earth (at some altitude) where the value of g will be equal to the value at the equator. So (A) is correct.
On the surface of the earth, the value of g is maximum at poles. So we will not get the same value at higher up (outside the earth).
10. 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
ANSWER: (a)
EXPLANATION: If the radius of the orbit = a, then velocity of an earth satellite v = √(GM/a) and the time period T = 2πa/v.
As we can see that both v and T are independent of the mass of the satellite but both depend on the radius of the orbit a.
11. The magnitude of gravitational potential energy of the moon-earth system is U with zero potential energy at infinite separation. the kinetic energy of the moon with respect to earth is K.
(a) U<K (b) U>K (d) U=K
ANSWER: (b)
EXPLANATION: Let the mass of the moon = m, velocity v and its distance from the earth = a.
v = √(GM/a), K.E. = K = ½mv² = ½m*GM/a
→GMm/a = 2K
The magnitude of gravitational potential energy of the moon
U = GMm/a =2K
Obviously U>K.
12. Figure (11-Q2) shows the elliptical path of a planet about the sun. The two shaded parts have equal area. If t₁ and t₂ be the time taken by the planet to go from a to b and from c to d respectively,
(a) t₁ < t₂
(b) t₁ = t₂
(c) t₁ > t₂
(d) insufficient information to deduce the relation between t₁ and t₂.
ANSWER: (b)
EXPLANATION: It is the second law of the planetary motion in elliptical orbit by Kepler that the radius vector from Sun to the planet sweeps out equal area in equal time.
13. A person sitting in a chair in a satellite feels weightless because
(a) the earth does not attract the object in a satellite
(b) the normal force by the chair on the person balances the earth's attraction
(c) the normal force is zero
(d) the person in a satellite is not accelerated
ANSWER: (c)
EXPLANATION: Only (c) is correct. The weight and the centrifugal force balance each other, hence the normal force is zero.
14. A body is suspended from a spring balance kept in a satellite. The reading of the balance is W₁ when the satellite goes in an orbit of radius R and is W₂ when it goes in an orbit of radius 2R.
(a) W₁ = W₂
(b) W₁ < W₂
(c) W₁ > W₂
(d) W₁ ≠ W₂
ANSWER: (a)
EXPLANATION: The weight of a body in a satellite in an orbit is zero whatever the radius of the orbit may be. If the spring balance shows a reading W₁ then it must be due to rotation of the satellite about its own axis. Assuming this rotation same in another orbit, W₁ = W₂. See the diagram below,
15. The kinetic energy needed to project a body of mass m from the earth's surface to infinity is
(a) ¼mgR
(b) ½mgR
(c) mgR
(d) 2mgR
ANSWER: (c)
EXPLANATION: The kinetic energy should be equal to the gravitational potential energy K=U = GMm/R.
Since g = GM/R² →GM =gR²
Hence, K = gR²*m/R =mgR
16. A particle is kept at rest at a distance R (earth's radius) above the earth's surface. The minimum speed with which it should be projected so that it does not return is
(a) √(GM/4R)
(b) √(GM/2R)
(c) √(GM/R)
(d) √(2GM/R)
ANSWER: (c)
EXPLANATION: The particle at rest is at 2R away from the earth's center. So, Escape velocity u = √(2GM/2R) = √(GM/R)
17. A satellite is orbiting the earth close to its surface. A particle is to be projected from the satellite to just escape from the earth. The escape speed from the earth is vₑ. It's speed with respect to the satellite
(a) will be less than vₑ
(b) will be more than vₑ
(c) will be equal to vₑ
(d) will depend on the direction of projection.
ANSWER: (d)
EXPLANATION: Its minimum escape speed will be vₑ when projected vertically upward, but if projected otherwise the speed will be greater. So the escape speed will depend on the direction of projection.
(a) 10 m/s² (b) 0.0027 m/s² (c) 6.4 m/s² (d) 5.0 m/s²
ANSWER: (b)
EXPLANATION: When moving in the orbit, the moon balances the force on it by the earth with its uniform circular motion in the orbit resulting in centrifugal force. When stopped for an instant, the force on the moon by the earth does not change, only balancing force is absent. So acceleration at that instant = Force /mass remains the same. Only in the orbit, it is the instantaneous centripetal acceleration.
2. The acceleration of the moon just before it strikes the earth in the previous question is
(a) 10 m/s² (b) 0.0027 m/s² (c) 6.4 m/s² (d) 5.0 m/s²
ANSWER: (c)
EXPLANATION: When the moon strikes the earth its center is R/4 above the surface of the earth. So the acceleration of the moon will be calculated at R+R/4 = 5R/4 away from the center of the earth. Since the acceleration is inversely proportional to the square of the distance, the acceleration of the moon at the time of strike
= g*{R/(5R/4)}² = g*(4/5)² = 16g/25 =160/25 =6.4 m/s².
3. Suppose the acceleration due to gravity at the earth's surface is 10 m/s² and at the surface of Mars it is 4.0 m/s². A 60 kg passenger goes from the earth to the Mars in a spaceship moving with a constant velocity. Neglect all other objects in the sky. Which part of the figure (11-Q1) best represents the weight (net gravitational force) of the passenger as a function of time.
(a) A (b) B (c) C (d) D
 |
| Figure for Q-3 |
ANSWER: (c)
EXPLANATION: Since the acceleration due to gravity varies inversely to the square of the distance hence the apparent weight (net gravitational force) of the passenger with respect to time will not be a straight line but a curve. In between the earth and the mars, there will be a point where the gravitational force due to both of the bodies will be equal and opposite, hence the apparent weight will zero but never negative. Out of the three curves in the figure, only curve C fulfills this condition.
4. Consider a planet in some solar system which has a mass double the mass of the earth and density equal to the average density of the earth. An object weighing W on the earth will weigh
(a) W (b) 2W (c) W/2 (d) 2¹/³ W at the planet.
ANSWER: (d)
EXPLANATION: Since the density is same the volume and hence the radius of the planet will be more than the radius of the earth. Let the radius of the earth be R and that of the planet be R'. If the density is ρ, then
4πR'³ρ/3 = 2*4πR³ρ/3
→R'³ =2R³
→R' = 2¹/³R
Acceleration due to gravity on this planet = G*2M/R'²
(M is mass of the earth)
=2GM/(2¹/³R)²
=2⁽¹⁻²/³⁾(GM/R²)
=2¹/³g
Hence the weight on the planet =2¹/³*mg =2¹/³W.
5. If the acceleration due to gravity at the surface of the earth is g, the work-done in slowly lifting a body of mass m from the earth's surface to a height R equal to the radius of the earth is
(a) ½mgR (b) 2mgR (c) mgR (d) ¼mgR
ANSWER: (a)
EXPLANATION: The distance of the body from the center of the earth at a height R from the earth's surface = 2R.
The acceleration due to gravity at distance x from the center of the earth ( for x>R) = GM/x²
Force on the body by the earth =mGM/x²
Since the body is lifted slowly, hence the moving force is ≈ mGM/x²
Hence the work done by the force in taking the body from R to 2R
= ∫m(GM/x²)dx
(limit of integration from R to 2R)
=mGM[-1/x]
=mGM[-1/2R+1/R]
=mGM/2R =½m(GM/R²)R = ½mgR
6. A person brings a mass of 1 kg from infinity to a point A. Initially, the mass was at rest but it moves at a speed of 2 m/s as it reaches A. The work done by the person on the mass is -3 J. The potential at A is
(a) -3 J/kg (b) -2 J/kg (c) -5 J/kg (d) None of these
ANSWER: (c)
EXPLANATION: Work done by the person = -3 J (Given)
Work done in increasing the velocity from rest to 2 km/s = K.E. of the mass at A =½mv² = ½*1*2² = 2 J
Let the potential at A be X J/kg,
So, the change in P.E. of the mass = the work done in slowly bringing the mass from infinity to A = mX
= 1*X = X J
So the total work done = Change in P.E + Change in K.E. = X+2 J, but it is given -3 J
so, X+2 = -3
→X = -3-2 = -5 J
7. Let V and E be the gravitational potential and the gravitational field at a distance r from the center of a uniform shell. Consider the following two statements:
(A) the plot of V against r is discontinuous.
(B) the plot of E against r is discontinuous.
(a) Both A and B are correct.
(b) A is correct but B is wrong.
(c) B is correct but A is wrong.
(d) Both A and B are wrong.
ANSWER: (c)
EXPLANATION: (A) is not correct because the gravitational potential inside the shell is constant =-GM/a, where a is the radius of the shell. As the distance increases from a, the potential gradually starts increasing till infinity to a value zero. So the curve is not discontinuous.
(B) is correct because the gravitational field inside the shell is zero and just as it goes outside the value jumps to GM/a² and then gradually decreases to zero at infinity. So the curve at r =a is discontinuous.
8. Let V and E be the gravitational potential and the gravitational field at a distance r from the center of a uniform solid sphere. Consider the following two statements:
(A) the plot of V against r is discontinuous.
(B) the plot of E against r is discontinuous.
(a) Both A and B are correct.
(b) A is correct but B is wrong.
(c) B is correct but A is wrong.
(d) Both A and B are wrong.
ANSWER: (d)
EXPLANATION: (A) is wrong because the value of V at the center is -3GM/2a and it increases continuously to -GM/a at the surface and from -GM/a at the surface to zero at the infinity. So it is continuous.
(B) is wrong because the value of E increases linearly from zero at the center to GM/a² at the surface, then decreases from GM/a² at the surface to zero at the infinity. So it is also continuous.
9. Take the effect of bulging of the earth and its rotation in the account. Consider the following statements:
(A) There are points outside the earth where the value of g is equal to its value at the equator.
(B) There are points outside the earth where the value of g is equal to its value at the poles.
(a) Both A and B are correct.
(b) A is correct but B is wrong.(c) B is correct but A is wrong.
(d) Both A and B are wrong.
ANSWER: (b)
EXPLANATION: As we move towards poles the value of g starts increasing. Another fact is that as we go up at a place the value of g starts decreasing. So at any point on the surface of the earth where the value of g is more than at the equator, there will be points outside the earth (at some altitude) where the value of g will be equal to the value at the equator. So (A) is correct.
On the surface of the earth, the value of g is maximum at poles. So we will not get the same value at higher up (outside the earth).
10. 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
ANSWER: (a)
EXPLANATION: If the radius of the orbit = a, then velocity of an earth satellite v = √(GM/a) and the time period T = 2πa/v.
As we can see that both v and T are independent of the mass of the satellite but both depend on the radius of the orbit a.
11. The magnitude of gravitational potential energy of the moon-earth system is U with zero potential energy at infinite separation. the kinetic energy of the moon with respect to earth is K.
(a) U<K (b) U>K (d) U=K
ANSWER: (b)
EXPLANATION: Let the mass of the moon = m, velocity v and its distance from the earth = a.
v = √(GM/a), K.E. = K = ½mv² = ½m*GM/a
→GMm/a = 2K
The magnitude of gravitational potential energy of the moon
U = GMm/a =2K
Obviously U>K.
12. Figure (11-Q2) shows the elliptical path of a planet about the sun. The two shaded parts have equal area. If t₁ and t₂ be the time taken by the planet to go from a to b and from c to d respectively,
(a) t₁ < t₂
(b) t₁ = t₂
(c) t₁ > t₂
(d) insufficient information to deduce the relation between t₁ and t₂.
 |
| Figure for Q-12 |
ANSWER: (b)
EXPLANATION: It is the second law of the planetary motion in elliptical orbit by Kepler that the radius vector from Sun to the planet sweeps out equal area in equal time.
13. A person sitting in a chair in a satellite feels weightless because
(a) the earth does not attract the object in a satellite
(b) the normal force by the chair on the person balances the earth's attraction
(c) the normal force is zero
(d) the person in a satellite is not accelerated
ANSWER: (c)
EXPLANATION: Only (c) is correct. The weight and the centrifugal force balance each other, hence the normal force is zero.
14. A body is suspended from a spring balance kept in a satellite. The reading of the balance is W₁ when the satellite goes in an orbit of radius R and is W₂ when it goes in an orbit of radius 2R.
(a) W₁ = W₂
(b) W₁ < W₂
(c) W₁ > W₂
(d) W₁ ≠ W₂
ANSWER: (a)
EXPLANATION: The weight of a body in a satellite in an orbit is zero whatever the radius of the orbit may be. If the spring balance shows a reading W₁ then it must be due to rotation of the satellite about its own axis. Assuming this rotation same in another orbit, W₁ = W₂. See the diagram below,
 |
| Reading of the spring balance for the weight of a body in a satellite in different orbits. |
15. The kinetic energy needed to project a body of mass m from the earth's surface to infinity is
(a) ¼mgR
(b) ½mgR
(c) mgR
(d) 2mgR
ANSWER: (c)
EXPLANATION: The kinetic energy should be equal to the gravitational potential energy K=U = GMm/R.
Since g = GM/R² →GM =gR²
Hence, K = gR²*m/R =mgR
16. A particle is kept at rest at a distance R (earth's radius) above the earth's surface. The minimum speed with which it should be projected so that it does not return is
(a) √(GM/4R)
(b) √(GM/2R)
(c) √(GM/R)
(d) √(2GM/R)
ANSWER: (c)
EXPLANATION: The particle at rest is at 2R away from the earth's center. So, Escape velocity u = √(2GM/2R) = √(GM/R)
17. A satellite is orbiting the earth close to its surface. A particle is to be projected from the satellite to just escape from the earth. The escape speed from the earth is vₑ. It's speed with respect to the satellite
(a) will be less than vₑ
(b) will be more than vₑ
(c) will be equal to vₑ
(d) will depend on the direction of projection.
ANSWER: (d)
EXPLANATION: Its minimum escape speed will be vₑ when projected vertically upward, but if projected otherwise the speed will be greater. So the escape speed will depend on the direction of projection.
===<<<O>>>===
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CHAPTER- 11 - Gravitation
Questions for Short Answers
CHAPTER- 10 - Rotational Mechanics
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OBJECTIVE - I
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EXERCISES - Q-1 TO Q-15
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EXERCISES Q-31 TO Q-45
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CHAPTER- 10 - Rotational Mechanics
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CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
HC Verma's Concepts of Physics, Chapter-8, WORK AND ENERGY
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ReplyDeleteSir last question????? I didn't get it
ReplyDeleteEscape velocity is the minimum velocity given to an object so that it will not return to the earth. For the velocity to be minimum, it should be projected vertically upward (radially away from the earth). If it is projected at some angle ß with the vertical and the velocity is V, then V.cosß should be equal to Ve. Thus V=Ve/cosß. Since cosß<1 hence V>Ve. So if it is projected in some other direction V will be more than Ve. It depends on the direction of projection.
DeleteSir in question 7th and 8th para 2nd is same but answer is different i.e continuous or discontinuous why?
ReplyDeleteDear student,
DeleteIn one problem the reference is uniform shell while in the other it is uniform solid sphere.