Physics | MHT CET 2017 | Discussion Page

The depth 'd' at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the earth's surface is

(R= radius of earth)

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

$d=R\left(\frac{n-1}{2n}\right)$

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

$d=R^2\left(\frac{n-1}{n}\right)$

Option C

Acceleration due to gravity at depth d is given as,

g'= $g\left(1-\frac{d}{R}\right)$ ...........(i)

Given, $g'=\frac{g}{n}$

Substituting the value of 'g' in Eq.(i) , we get

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

$\Rightarrow$ $\frac{1}{n}=1-\frac{d}{R}$

$\therefore$ $\frac{d}{R}=1-\frac{1}{n}=\frac{n-1}{n}$

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

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