51. The ratio of escape velocity at earth (v_e) to the escape velocity at a planet (v_p) whose radius and mean density are twice as that of earth is :

1. 1. 1 : 2

   
   
   

   2. 1 : 2 ​√2

   
   
   

   3. 1 : 4 ​

   
   
   

   4. 1 : 2

   
   
   

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1. View Hint View Answer Discuss in Forum

   As we know, escape velocity,
   
   

   ⇒	Vc	=	Rc	√ρc / 2ρc
   	Vp		2Rc

   
   

   ∴ Ratio	Vc	= 1 : 2 √2
   	Vp

   Correct Option: B

   As we know, escape velocity,
   
   

   ⇒	Vc	=	Rc	√ρc / 2ρc
   	Vp		2Rc

   
   

   ∴ Ratio	Vc	= 1 : 2 √2
   	Vp

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52. With what velocity should a particle be projected so that its height becomes equal to radius of earth?​​​

1. 1. 		GM		1 / 2
      		R

   
   
   

   2. 		8GM		1 / 2
      		R

   
   
   

   3. 		2GM		1 / 2
      		R

   
   
   

   4. 		4GM		1 / 2
      		R

   
   
   

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1. View Hint View Answer Discuss in Forum

   From conservation of energy
   

   	1	mu2 -	GMm	=	1	m × (0)2 -	GMm
   	2		R		2		R + R

   
   

   ⇒ u2 =	2GM	-	2GM	=	GM
   	R		2R		R

   
   
   

   Correct Option: A

   From conservation of energy
   

   	1	mu2 -	GMm	=	1	m × (0)2 -	GMm
   	2		R		2		R + R

   
   

   ⇒ u2 =	2GM	-	2GM	=	GM
   	R		2R		R

   
   
   

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53. ​The potential energy of a satellite, having mass m and rotating at a height of 6.4 × 10⁶ m from the earth surface, is​​​

1. 1. – mgR_e ​

   
   
   

   2. – 0.67 mgR_e

   
   
   

   3. ​– 0.5 mgR_e

   
   
   

   4. – 0.33 mgR_e

   
   
   

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1. View Hint View Answer Discuss in Forum

   Mass of the satellite = m and height of satellite from earth (h) = 6.4 × 10⁶ m. ​
   We know that gravitational potential energy of the satellite at height
   

   h = -	G Me m	= -	gRe2 m
   	(Re + h)		2Re

   
   

   = -	gRem	= -0.5 mgRe
   	2

   
   (where, GM_e = gR_e​² and h = R_e)

   Correct Option: C

   Mass of the satellite = m and height of satellite from earth (h) = 6.4 × 10⁶ m. ​
   We know that gravitational potential energy of the satellite at height
   

   h = -	G Me m	= -	gRe2 m
   	(Re + h)		2Re

   
   

   = -	gRem	= -0.5 mgRe
   	2

   
   (where, GM_e = gR_e​² and h = R_e)

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54. Assuming the radius of the earth as R, the change in gravitational potential energy of a body of mass m, when it is taken from the earth's surface to a height 3R above its surface, is

1. 1. 3 mg R

   
   
   

   2. 	3	mgR
      	4

   
   
   

   3. 1 mgR

   
   
   

   4. 	3	mgR
      	2

   
   
   

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1. View Hint View Answer Discuss in Forum

   Gravitational potential energy (GPE) on the surface of earth,
   

   E1 = -	G M m
   	R

   
   

   GPE at 3R, E2 = -	G M m	= -	G M m
   	(R + 3R)		4R

   
   

   ∴ Change in GPE = E2 - E1 = -	GMm	+	GMm	=	3GMm
   	4R		R		4R

   
   

   =	3g R2m		∵ g =	GM
   	4R			R2

   
   

   =	3	mg R
   	4

   Correct Option: B

   Gravitational potential energy (GPE) on the surface of earth,
   

   E1 = -	G M m
   	R

   
   

   GPE at 3R, E2 = -	G M m	= -	G M m
   	(R + 3R)		4R

   
   

   ∴ Change in GPE = E2 - E1 = -	GMm	+	GMm	=	3GMm
   	4R		R		4R

   
   

   =	3g R2m		∵ g =	GM
   	4R			R2

   
   

   =	3	mg R
   	4

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