The depth at which the value of acceleration due to gravity becomes (1/n) times the value at the surface of the earth is ( R = radius of the earth) (a) (R(n-1))/(n) (b) (R(n+1))/(n) (c) (Rn)/((n-1)) (d) R/n

Answer: (a). At depth d: gd = g(1 - d/R). 1 - d/R = 1/n d = R(1 - 1/n) = (R(n-1))/(n) So the answer is (a) (R(n-1))/(n).

The depth at which the value of acceleration due to gravity becomes… | MHT CET 2025 (25 Apr, Shift 2)

Q19MHT CET 2025 · 25 APR · SHIFT 2Easy

Acceleration Due to Gravity

The depth at which the value of acceleration due to gravity becomes $\left(\frac{1}{n}\right)$ times the value at the surface of the earth is

( $R =$ radius of the earth)

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Answer: (a)

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