The depth d below the surface of the earth wh | vedclass

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Question

(A) The acceleration due to gravity at a depth $d$ below the surface of the earth is given by the formula: $g' = g\left(1 - \frac{d}{R}\right)$.Accord

Vedclass · Accepted Answer

$R\left(\frac{n-1}{n}\right)$

Vedclass · Answer

(A) The acceleration due to gravity at a depth $d$ below the surface of the earth is given by the formula: $g' = g\left(1 - \frac{d}{R}\right)$.According to the problem,the value of acceleration due to gravity at depth $d$ is $\frac{1}{n}$ times the value at the surface,so $g' = \frac{g}{n}$.Substituting this into the formula: $\frac{g}{n} = g\left(1 - \frac{d}{R}\right)$.Dividing both sides by $g$: $\frac{1}{n} = 1 - \frac{d}{R}$.Rearranging the terms to solve for $d$: $\frac{d}{R} = 1 - \frac{1}{n} = \frac{n-1}{n}$.Therefore,$d = R\left(\frac{n-1}{n}\right)$.

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 $d$ below the surface of the earth where the value of acceleration due to gravity becomes $\left(\frac{1}{n}\right)$ times the value at the surface of the earth is ($R$ = radius of the earth).

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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

If the radius of the Earth were to shrink by half,its mass remaining the same,what would be the weight of an object on its surface?

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