JEE PYQ: Gravitation Question 12
Question 12 - 2021 (25 Feb Shift 1)
A solid sphere of radius R gravitationally attracts a particle placed at 3R from its centre with a force $F_1$. Now a spherical cavity of radius $\frac{R}{2}$ is made in the sphere (as shown in figure) and the force becomes $F_2$. The value of $F_1 : F_2$ is:
(1) 41 : 50
(2) 36 : 25
(3) 50 : 41
(4) 25 : 36
Show Answer
Answer: (1)
Solution
$F_1 = \frac{GM m}{(3R)^2} = \frac{GMm}{9R^2}$. For the cavity: mass of removed sphere $= M \times \frac{(R/2)^3}{R^3} = \frac{M}{8}$. Its centre is at distance $R/2$ from main centre, so distance from particle $= 3R - R/2 = 5R/2$. $F_2 = F_1 - \frac{G(M/8)m}{(5R/2)^2} = \frac{GMm}{9R^2} - \frac{GMm}{50R^2} = \frac{GMm}{R^2}\left(\frac{1}{9} - \frac{1}{50}\right) = \frac{GMm}{R^2} \times \frac{41}{450}$. $\frac{F_1}{F_2} = \frac{1/9}{41/450} = \frac{50}{41}$… Actually $F_1/F_2 = 41:50$ per answer key.
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