A solid sphere of radius R gravitationally at | vedclass

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(C) Let the initial mass of the solid sphere be $M$. The density of the sphere is $\rho = \frac{M}{\frac{4}{3}\pi R^3}$.The force $F_{1}$ on a particl

Vedclass · Accepted Answer

$50: 41$

Vedclass · Answer

(C) Let the initial mass of the solid sphere be $M$. The density of the sphere is $\rho = \frac{M}{\frac{4}{3}\pi R^3}$.The force $F_{1}$ on a particle of mass $m$ at a distance $3R$ from the centre is given by:$F_{1} = \frac{GMm}{(3R)^2} = \frac{GMm}{9R^2}$.When a spherical cavity of radius $r = \frac{R}{2}$ is made,the mass of the removed portion is $M' = \rho \cdot \frac{4}{3}\pi (\frac{R}{2})^3 = M \cdot (\frac{1}{2})^3 = \frac{M}{8}$.The centre of the cavity is at a distance of $\frac{R}{2}$ from the centre of the sphere. The particle is at a distance $3R$ from the centre of the sphere,so it is at a distance $3R - \frac{R}{2} = \frac{5R}{2}$ from the centre of the cavity.The new force $F_{2}$ is the force due to the original sphere minus the force due to the removed portion:$F_{2} = \frac{GMm}{(3R)^2} - \frac{G(M/8)m}{(5R/2)^2} = \frac{GMm}{9R^2} - \frac{GMm}{8 \cdot \frac{25R^2}{4}} = \frac{GMm}{R^2} (\frac{1}{9} - \frac{1}{50})$.$F_{2} = \frac{GMm}{R^2} (\frac{50 - 9}{450}) = \frac{41}{450} \frac{GMm}{R^2}$.Therefore,the ratio $F_{1} : F_{2}$ is:$\frac{F_{1}}{F_{2}} = \frac{GMm/9R^2}{41GMm/450R^2} = \frac{1}{9} \cdot \frac{450}{41} = \frac{50}{41}$.Thus,the ratio is $50:41$.

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