A spherical cavity is made in a lead sphere of radius R such than its surface touches the outside surface of the lead sphere and passes through its centre. The mass of the lead sphere before hollowing was M. The force of attraction that this sphere would exert on a particle of mass m, which lies at a distance d from the centre of the lead sphere on the straight line joining the centres of the sphere and the centre of the cavity as shown in the figure is

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$\frac{G M m}{2 R^{2}} \hat{i}$

$- \frac{G M m}{2 R^{2}} \hat{i}$

$- \frac{G M m d}{2 R^{3}} \hat{i}$

$\frac{G M m (\frac{R}{2} - d)}{R^{3}} \hat{i}$

answer is B.

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Detailed Solution

Force calculation of gravitational field intensity inside the cavity.

$G r a v i t a t i o n a l f i e l d i n t e n s i t y w i t h o u t c a v i t y, E_{1} = \frac{G M d}{R^{3}} (- \hat{i})$
Radius of cavity $R' = \frac{R}{2}, r = (\frac{R}{2} - d)$ i.e., inside the cavity the gravitational field intensity,
$I n t e n s i t y d u e t o r e m a i n i n g p a r t E_{2} = \frac{G M' r}{R'^{3}}$
$E_{2} = \frac{G M / 8 (\frac{R}{2} - d) (\hat{i})}{{(\frac{R}{2})}^{3}} E_{2} = \frac{G M}{R^{3}} (\frac{R}{2} - d) \hat{i}$
Gravitational field intensity due to cavity portion of radius $(\frac{R}{2} - d)$
$n e t i n t e n s i t y E = E_{1} - E_{2} E = \frac{G M d}{R^{3}} (- \hat{i}) + \frac{G M}{R^{3}} (\frac{R}{2} - d) \hat{i} h e r e u n i t v e c t o r \hat{i} a r e o p p o s i t e l y d i r e c t e d h e n c e t a k e n a d d i t i o n E = \frac{G M}{R^{3}} (d + \frac{R}{2} - d) (- \hat{i}) E = \frac{G M}{R^{3}} (\frac{R}{2}) (- \hat{i}) f o r c e F = E \times m$
Force on m, $F = - \frac{G M m}{2 R^{2}} \hat{i}$