The depth d at which the value of acceleratio | vedclass

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(A) The acceleration due to gravity at a depth $d$ below the surface of the earth is given by $g_d = g_s \left(1 - \frac{d}{R}\right)$,where $g_s$ is

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The acceleration due to gravity at a depth $d$ below the surface of the earth is given by $g_d = g_s \left(1 - \frac{d}{R}\right)$,where $g_s$ is the acceleration due to gravity at the surface.Given that $g_d = \frac{g_s}{n}$,we substitute this into the equation:$\frac{g_s}{n} = g_s \left(1 - \frac{d}{R}\right)$Dividing both sides by $g_s$:$\frac{1}{n} = 1 - \frac{d}{R}$Rearranging the terms to solve for $d$:$\frac{d}{R} = 1 - \frac{1}{n}$$\frac{d}{R} = \frac{n-1}{n}$$d = \frac{R(n-1)}{n}$Thus,the correct option is $A$.

The depth $d$ at which 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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