A cavity of radius R/2 is made inside a solid sphere of radius R. The centre of the cavity is located at a distance R/2 from the centre of the sphere. The gravitational force on a particle of mass ‘m’ at a distance R/2 from the centre of the sphere on the line joining both the centers of sphere and cavity is (opposite to the centre of cavity).[Here g = GM/ $R^{2}$ , where M is the mass of the sphere]

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$\frac{m g}{2}$

$\frac{3}{8} m g$

$m g$

Zero

answer is B.

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Detailed Solution

The case can be considered as a sphere of density ρ and another sphere of density −ρ of the exact dimensions and position as the cavity

Attractive force due to the sphere of density

$ρ = \frac{G (ρ \frac{4}{3} π (R / 2)^{3}) m}{(R / 2)^{2}} = \frac{mg}{2}$ $(since M = ρ \frac{4}{3} {πR}^{3})$

Attractive force due to the sphere of density $- ρ = - \frac{G (ρπ (R / 2)^{3})}{R^{2}} = - \frac{mg}{8}$

Hence net force

$\frac{mg}{2} - \frac{mg}{8} = \frac{3 mg}{8}$

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