First slide

Variation of g

Question

What is the depth at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value that at the surface of earth? (radius of earth = R)

Very Easy

A

$\frac{R}{n}$
B

$\frac{R}{n^{2}}$
C

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

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

Solution

Value of acceleration due to gravity at depth d,
$\begin{array}{l} g^{'} = g (1 - \frac{d}{R}) \\ \frac{g}{n} = g (1 - \frac{d}{R}) \\ 1 - \frac{d}{R} = \frac{1}{n} \\ \frac{d}{R} = 1 - \frac{1}{n} = (\frac{n - 1}{n}) \\ d = R (\frac{n - 1}{n}) \end{array}$

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