Q#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)
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.
Q#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)
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.
Q#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)
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.
Q#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)
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' = (2)1/3 R
Acceleration due to gravity on this planet = G(2M)/R'²
(M is mass of the earth)
= 2GM/[(2)1/3R]²
= (GM/R²)
= 21/3g
Hence the weight on the planet = 21/3mg = 21/3W.
Q#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)
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
Q#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)
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)(2)² = 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
Q#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)
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.
Q#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)
(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.
Q#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)
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).
(a) 10 m/s² (b) 0.0027 m/s² (c) 6.4 m/s² (d) 5.0 m/s²
Answer: (c)
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.
Q#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)
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.
Q#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)
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' = (2)1/3 R
Acceleration due to gravity on this planet = G(2M)/R'²
(M is mass of the earth)
= 2GM/[(2)1/3R]²
= (GM/R²)
= 21/3g
Hence the weight on the planet = 21/3mg = 21/3W.
Q#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)
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
Q#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)
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)(2)² = 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
Q#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)
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.
Q#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)
(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.
Q#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)
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).
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