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(C) The gravitational acceleration $g$ at a distance $r$ from the centre of a uniform solid sphere of mass $M$ and radius $R$ (where $r \ge R$) is giv

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$2 R$

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(C) The gravitational acceleration $g$ at a distance $r$ from the centre of a uniform solid sphere of mass $M$ and radius $R$ (where $r \ge R$) is given by $g = \frac{GM}{r^2}$.On the surface,$r = R$,so $a_o = \frac{GM}{R^2}$.We want to find the distance $r$ where the acceleration becomes $a = \frac{a_o}{4}$.Substituting the expressions,we get $\frac{GM}{r^2} = \frac{1}{4} \left( \frac{GM}{R^2} \right)$.Canceling $GM$ from both sides,we get $\frac{1}{r^2} = \frac{1}{4R^2}$.Taking the square root of both sides,$r^2 = 4R^2$,which gives $r = 2R$.Thus,the distance from the centre of the sphere is $2R$.

$A$ uniform solid sphere of radius $R$ produces a gravitational acceleration of $a_o$ on its surface. The distance of the point from the centre of the sphere where the gravitational acceleration becomes $\frac{a_o}{4}$ is,

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