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Figure shows a spherical cavity inside a lead sphere. The surface of the cavity passes through the centre of the sphere and touches the right side of the sphere. The mass of the sphere before hollowing was M. The gravitational force does the hollowed out lead sphere attract a particle of mass m that lies at a distance d from the centre of the lead sphere on the straight line connecting the centres of the spheres and of the cavity is $\frac{G M m}{d^{2}} (1 - \frac{1}{x {(1 - \frac{R}{2 d})}^{2}}) .$ The value of ‘x’ is

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answer is 8.

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Detailed Solution

Mass of the sphere removed $M' = \frac{M}{8}$

$F = \frac{G M m}{d^{2}} - \frac{G M' m}{{(d - R / 2)}^{2}}$

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