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(C) The gravitational force on mass $m$ due to the complete sphere is given by:$F_1 = \frac{GMm}{(2R)^2} = \frac{GMm}{4R^2} \quad ...(1)$When a spheri

Vedclass · Accepted Answer

$12: 11$

Vedclass · Answer

(C) The gravitational force on mass $m$ due to the complete sphere is given by:$F_1 = \frac{GMm}{(2R)^2} = \frac{GMm}{4R^2} \quad ...(1)$When a spherical part of radius $r = R/3$ is removed,its mass $M'$ is calculated by considering the density $ho$ of the sphere:$M' = ho \cdot V' = \left( \frac{M}{\frac{4}{3}\pi R^3} ight) \cdot \left( \frac{4}{3}\pi (R/3)^3 ight) = \frac{M}{27}$The center of the removed sphere is at a distance $d = R - R/3 = 2R/3$ from the center $O$. The distance of this center from the point mass $m$ is $2R - 2R/3 = 4R/3$.The force $F_2$ exerted by the remaining part is the original force minus the force that would have been exerted by the removed part:$F_2 = F_1 - F_{	ext{removed}} = \frac{GMm}{4R^2} - \frac{G(M/27)m}{(4R/3)^2}$$F_2 = \frac{GMm}{4R^2} - \frac{GMm}{27 \cdot (16R^2/9)} = \frac{GMm}{4R^2} - \frac{GMm}{48R^2}$$F_2 = \frac{GMm}{R^2} \left( \frac{1}{4} - \frac{1}{48} ight) = \frac{GMm}{R^2} \left( \frac{12-1}{48} ight) = \frac{11}{48} \frac{GMm}{R^2}$Now,the ratio $F_1 : F_2$ is:$\frac{F_1}{F_2} = \frac{GMm / 4R^2}{11GMm / 48R^2} = \frac{1}{4} \cdot \frac{48}{11} = \frac{12}{11}$Thus,$F_1 : F_2 = 12 : 11$.

$\text{LIST-I}$	$\text{LIST-II}$
$A$. Gravitational constant	$I$. $[LT^{-2}]$
$B$. Gravitational potential energy	$II$. $[L^2 T^{-2}]$
$C$. Gravitational potential	$III$. $[ML^2 T^{-2}]$
$D$. Acceleration due to gravity	$IV$. $[M^{-1} L^3 T^{-2}]$

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