Q106 / 150 · MHT CET 2021 (20 Sep Shift 2) · Physics

The depth ' $d$ ' below the surface of the earth where the value of acceleration due to gravity becomes $\left(\frac{1}{n}\right)$ times the value at the surface of the earth is $(\mathrm{R}$ = radius of the earth)

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

Solution

Inside the earth, $g^{\prime}=g\left(1-\frac{d}{R}\right)$ $$ \frac{\mathrm{g}}{\mathrm{n}}=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right) $$ Or $\mathrm{d}=\mathrm{R}\left(\frac{\mathrm{n}-1}{\mathrm{n}}\right)$

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

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