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Question

Physics Question on Newtons law of gravitation

The depth dd at which the value of acceleration due to gravity becomes 1n\frac {1} { n} time the value at the surface is (R be the radius of the earth)

A

R/n

B

R/n2R/n^2

C

R(n−1)nR\frac{(n-1)}{n}

D

Rn(n−1)\frac{Rn}{(n-1)}

Answer

R(n−1)nR\frac{(n-1)}{n}

Explanation

Solution

The correct option is(C): R(n−1)nR\frac{(n-1)}{n}.

At depth d from earth's surface,
g′=g(1−dR)g'=g\left(1-\frac{d}{R} \right)
Given g′=gng'=\frac{g}{n}
So, gn=g(1−dR)\frac{g}{n}=g\left(1-\frac{d}{R}\right)
or 1n=1−dR\frac{1}{n}=1-\frac{d}{R} or dR=1−1n\frac{d}{R}=1-\frac{1}{n}
=(n−1)Rn=\frac{(n-1)R}{n}
∴\therefore d=(n−1)Rnd=\frac{(n-1)R}{n}