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 $M$ . With what 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.

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$\frac{G M m}{d^{3}} [1 - \frac{1}{8 {(1 - \frac{R}{2 d})}^{2}}]$

$\frac{G M m}{d^{2}} [1 - \frac{1}{8 {(1 - \frac{R}{2 d})}^{2}}]$

$\frac{G M m}{d^{2}} [\frac{1}{8 {(1 - \frac{R}{2 d})}^{2}}]$

$\frac{G M m}{d} [1 - \frac{1}{{(1 - \frac{R}{2 d})}^{2}}]$

answer is B.

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Detailed Solution

Radius of hollow sphere is $\frac{R}{2}$ , so mass in this hollow portion would had been, $\frac{M}{8}$ .

Now, net force on $m$ due to whole sphere $=$ force due to remaining mass $+$ force due to cavity mass.

$∴$ Force due to remaining mass $=$ force due to whole sphere $-$ force due to cavify mass

$= \frac{G M m}{d^{2}} - \frac{G M m}{8 (d - R / 2)^{2}}$

$= \frac{G M m}{d^{2}} [1 - \frac{1}{8 {(1 - \frac{R}{2 d})}^{2}}]$

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