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Question

(A) The acceleration due to gravity at a height $h$ above the earth's surface is given by the formula:$g^{\prime} = g \left( \frac{R}{R+h} \right)^{2}

Vedclass · Accepted Answer

$(n+1)^{2}$

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(A) The acceleration due to gravity at a height $h$ above the earth's surface is given by the formula:$g^{\prime} = g \left( \frac{R}{R+h} \right)^{2}$Given that the height $h = nR$,we substitute this into the equation:$g^{\prime} = g \left( \frac{R}{R+nR} \right)^{2}$$g^{\prime} = g \left( \frac{R}{R(1+n)} \right)^{2}$$g^{\prime} = g \left( \frac{1}{1+n} \right)^{2}$Now,we find the ratio of the acceleration due to gravity on the surface $(g)$ to that at the altitude $(g^{\prime})$:$\frac{g}{g^{\prime}} = \frac{g}{g \left( \frac{1}{1+n} \right)^{2}}$$\frac{g}{g^{\prime}} = (1+n)^{2}$Therefore,the ratio is $(n+1)^{2}$.

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

$A$ body is taken to a height of $n R$ from the surface of the earth. The ratio of the acceleration due to gravity on the surface to that at the altitude is

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Assertion $(A)$: $A$ particle of mass $m$ dropped into a hole made along the diameter of the Earth from one end to the other possesses simple harmonic motion.
Reason $(R)$: Gravitational force between any two particles is inversely proportional to the square of the distance between them.

What effect occurs on the frequency of a pendulum if it is taken from the earth's surface to deep into a mine?

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

The depth $d$ at which the value of acceleration due to gravity becomes $\frac{1}{n-1}$ times the value at the earth's surface is ($R=$ radius of the earth).

The height of the point vertically above the earth's surface,at which the acceleration due to gravity becomes $1\%$ of its value at the surface is (Radius of the earth $= R$). (in $,R$)

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