Question

A solid sphere of mass M and radius R is surrounding by a spherical shell of same mass M and radius 2R as shown. A small particle of mass m is released from rest from a height (h <<R) above the shell there is a hole in the shell. 

(i) in what time will it enter the hole at A 

(A) [(2√hR²) / (GM)] 

(B) √[(2hR²) / (GM)] 

(C) √[(hR²) / (GM)] 

(D) None of these

(ii) What time will it take to move from A to B ? 

(A) = [R² / (√GMh)] 

(B) > [R² / (√GMh)] 

(C) < [R² / (√GMh)] 

(D) None of these 

(iii) with what approximate speed will it collide at B? 

(A) √[(2GM) / R] 

(B) √[(3GM) / (2R)] 

(C) √[(GM) / (2R)] 

(D) √(GM / R)

Answer 1

Correct option: (1) (A) [(2√hR²) / (GM)], (2) (C) < [R² / (√GMh)], (3) (D) √(GM / R)

Explanation:

Acceleration due to gravity near surface of shell can be assumed to be uniform.

g = [{G(2M)} / (2R²)] = [(GM) / (2R²)]

Also    h = (1/2)gt² hence t = √(2h / g)

t = √[(2h × 2R²) / (GM)] = 2√(hR² / GM)

 ∴ option A is answer

(2) (C) < [R² / (√GMh)]

∵ (1/2)mV_A² = mgh

V_A² = 2gh

V_A = √(2gh) = √[2 × {(GM) / (2R²)} × h]

= √[(GMh) / R²]

From A to B field due to shell is zero but field due to sphere is not zero.

hence t_AB < (R / V_A) < [R² / √(GMh)]

∴ option C is answer

(3) (D) √(GM / R)

K_A = 0 potential between A & B due to shell from energy conservation,

K_A + U_A = K_B + U_B 

K_B = U_A – U_B --------- as K_A = 0

∴  (1/2)mV_B² = m(V_A – V_B)

V_B = √[2(V_A – V_B)]

= √(GM / R)

∴ option D is answer