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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 $2R$.

Column-$I$	Column-$II$
$(1)$ Maximum value of acceleration due to gravity $g$	$(a)$ At the center of the Earth
$(2)$ Minimum value of acceleration due to gravity $g$	$(b)$ At the poles
$(3)$ Zero value of acceleration due to gravity $g$	$(c)$ At the equator

$A$ 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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Match Column-$I$ with Column-$II$.

Column-$I$ Column-$II$

$(1)$ Maximum value of acceleration due to gravity $g$ $(a)$ At the center of the Earth

$(2)$ Minimum value of acceleration due to gravity $g$ $(b)$ At the poles

$(3)$ Zero value of acceleration due to gravity $g$ $(c)$ At the equator

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