Engineering
Physics
Equilibrium
Question

Consider two solid spherical asteroid of uniform density of mass M and radius R. In one asteroid a tunnel of very small size of depth R is bored to the centre and in other asteroid a spherical cavity of radius R/2 is made as shown in the figure. Now, identical particles of mass m dropped into the cavities of both asteroids from the top most point P. If force experienced by particle is FI and FII respectively in cavities of asteroids I and II, when they are x distance away from the centre of asteroids. If the time taken by particles to reach the centre of asteroids is TI and TII respectively then

the ratio of TITII is equal to π2

the ratio of FIFII is equal to xR

the ratio of FIFII is equal to 2xR

the ratio of TITII is equal to π4

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Solution
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Asteroid-I

Field E = GMR3x

FIFII=2xR

T1=2π4R3GM

Asteroid-II

Field FII=GMR3(R2)

R=12GMR3(R2)TII2

TII=2R3GM

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