11. Assuming earth to be a sphere of uniform density, what is the value of ‘g’ in a mine 100 km below the earth’s surface? (Given, R = 6400 km)
    

1. 1. 9.65 m / s²
      

   
   
   

   2. 7.65 m / s²
      

   
   
   

   3. 5.06 m / s²
      

   
   
   

   4. 3.10 m / s²

   
   
   

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1. View Hint View Answer Discuss in Forum

   We know that effective gravity g' at depth  below earth surface is given by
   

   g' = g		1 -	d
   			R

   
   Here, d = 100 km, R = 6400 km,
   

   ∴ g' = 9.8		1 -	100		= 9.65 m / s2
   			6400

   Correct Option: A

   We know that effective gravity g' at depth  below earth surface is given by
   

   g' = g		1 -	d
   			R

   
   Here, d = 100 km, R = 6400 km,
   

   ∴ g' = 9.8		1 -	100		= 9.65 m / s2
   			6400

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12. ​The acceleration due to gravity on the planet A is 9 times the acceleration due to gravity on planet B. A man jumps to a height of 2m on the surface of A. What is the height of jump by the same person on the planet B?​​
    

1. 1. 	2	m
      	3

   
   
   

   2. 	2	m
      	9

   
   
   

   3. 18 m
      

   
   
   

   4. 6 m

   
   
   

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1. View Hint View Answer Discuss in Forum

   Applying conservation of total mechanical energy principle
   

   	1	mv2 = mgAhA = mgBhB
   	2

   
   ⇒ g_A h_A = g_B h_B
   

   ⇒ hB =		gA		hA = 9 × 2 = 18 m
   		gB

   
   

   Correct Option: C

   Applying conservation of total mechanical energy principle
   

   	1	mv2 = mgAhA = mgBhB
   	2

   
   ⇒ g_A h_A = g_B h_B
   

   ⇒ hB =		gA		hA = 9 × 2 = 18 m
   		gB

   
   

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13. The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is R, the radius of the planet would be​​
    

1. 1. (1 / 2) R

   
   
   

   2. 2 R

   
   
   

   3. ​4 R

   
   
   

   4. (1/4) R

   
   
   

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1. View Hint View Answer Discuss in Forum

   g =	GM	.Also , M = d ×	4	πR3
   	R2		3

   
   

   ∴ g = G	4	dπR
   	3

   
   

   At the surface of planet, gp =	4	G(2d)πR' ,
   	3

   
   

   At the surface of the earth ge =	4	GdπR
   	3

   
   g_e = g_p ⇒ dR = 2d R'  ⇒ R' = R/2

   Correct Option: A

   g =	GM	.Also , M = d ×	4	πR3
   	R2		3

   
   

   ∴ g = G	4	dπR
   	3

   
   

   At the surface of planet, gp =	4	G(2d)πR' ,
   	3

   
   

   At the surface of the earth ge =	4	GdπR
   	3

   
   g_e = g_p ⇒ dR = 2d R'  ⇒ R' = R/2

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14. Imagine a new planet having the same density as that of earth but it is 3 times bigger than the earth in size. If the acceleration due to gravity on the surface of earth is g and that on the surface of the new planet is g’, then
    

1. 1. g’ = (g / 9)

   
   
   

   2. g’ = 27g

   
   
   

   3. g’= 9g

   
   
   

   4. g’= 3g

   
   
   

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1. View Hint View Answer Discuss in Forum

   We know that
   

   g =

   GM

   =

   G

   4

   πR3

   ρ

   =

   4

   πGRρ

   3

   R2

   R2

   3

   
   

   	g'	=	R'	=	3R	= 3
   	g		R		R

   
   ∴ g' = 3g

   Correct Option: D

   We know that
   

   g =

   GM

   =

   G

   4

   πR3

   ρ

   =

   4

   πGRρ

   3

   R2

   R2

   3

   
   

   	g'	=	R'	=	3R	= 3
   	g		R		R

   
   ∴ g' = 3g

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15. The distance of two planets from the sun are 10¹³ and 10¹² metres respectively. The ratio of time periods of these two planets is​

1. 1. 	1
      	√10

   
   
   

   2. 100
      

   
   
   

   3. 10 √10
      

   
   
   

   4. √10

   
   
   

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1. View Hint View Answer Discuss in Forum

   T² ∝ R³ (Kepler's law)
   

   	T12	=			1013		3	⇒	T1	= 10 √10
   	T22				1012				T2

   Correct Option: C

   T² ∝ R³ (Kepler's law)
   

   	T12	=			1013		3	⇒	T1	= 10 √10
   	T22				1012				T2

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