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Home » The depth ′ d ′ at which the value of acceleration due to gravity becomes 1/n times the value at the earth’s surface is ( R = radius of earth)
The depth ′ d ′ at which the value of acceleration due to gravity becomes 1/n times the value at the earth’s surface is ( R = radius of earth)
Physics | Physics
The depth ′ d ′ at which the value of acceleration due to gravity becomes 1/n times the value at the earth’s surface is ( R = radius of earth)

Solution: The correct answer is B.

The following is the expression for acceleration due to gravity at a depth of d below the earth’s surface.

g'={{g}_{s}}\left( 1-\frac{d}{R} \right)

We\,have:\,g'=\frac{{{g}_{s}}}{n}

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

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

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

i

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