First slide

Variation of g

Question

The depth d at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the surface, is [R =radius of the earth]

Moderate

A

$\frac{R}{n}$
B

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

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

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

Solution

$g^{'} = g (1 - \frac{d}{R}) \Rightarrow \frac{g}{n} = g (1 - \frac{d}{R})$
$\Rightarrow \frac{d}{R} = 1 - \frac{1}{n} \Rightarrow d = (\frac{n - 1}{n}) R$

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