Depth at which value of acceleration due to gravity becomes frac{1}{n} times value that at

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7.Gravitation

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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 $=R$ )

A

$\frac{R}{n}$

B

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

C

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

D

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

(NEET-2020)

Solution

Value of acceleration due to gravity at depth d,

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

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

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

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

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

Standard 11

Physics

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