Q133 / 150 · MHT CET 2025 (26 Apr Shift 2) · Physics

The depth at which acceleration due to gravity becomes $\frac{\mathrm{g}}{\mathrm{n}}$ is ( $\mathrm{R}=$ radius of earth, $\mathrm{g}=$ acceleration due to gravity) ( $\mathrm{n}=$ integer)

  • Correct answer
    A
    $\frac{R(n-1)}{n}$
  • B
    $\frac{\mathrm{R}(\mathrm{n}+1)}{\mathrm{n}}$
  • C
    $\frac{\mathrm{R}(\mathrm{n}-1)^2}{\mathrm{n}}$
  • D
    $\frac{\mathrm{R}(\mathrm{n}+1)^2}{\mathrm{n}}$

Solution

The acceleration due to gravity at depth d below Earth's surface is given by $g' = g\left(1 - \frac{d}{R}\right)$.

Since $g' = \frac{g}{n}$, substitution yields $\frac{g}{n} = g\left(1 - \frac{d}{R}\right)$.

Dividing both sides by g and simplifying produces $\frac{d}{R} = 1 - \frac{1}{n} = \frac{n-1}{n}$.

The depth is therefore $d = R\cdot\frac{n-1}{n}$, which corresponds to option A.

Hence, the correct answer is option: 'A'.

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