Q114 / 150 · MHT CET 2024 (03 May Shift 2) · Physics

The height at which the weight of the body becomes $\frac{1^{\text {th }}}{16}$ of its weight on the surface of the earth of radius ' $R$ ' is

  • A
    2 R
  • Correct answer
    B
    3 R
  • C
    4 R
  • D
    5 R

Solution

At height $h=\frac{R}{n}$ the value of acceleration due to gravity is given by, $\begin{aligned} \quad \frac{\mathrm{g}_{\mathrm{h}}}{\mathrm{~g}} & =\left(\frac{\mathrm{n}}{\mathrm{n}+1}\right)^2=\frac{1}{16} \Rightarrow \frac{\mathrm{n}}{\mathrm{n}+1}=\frac{1}{4} \\ \therefore \quad \mathrm{n} & =\frac{1}{3} \\ \mathrm{~h} & =\frac{\mathrm{R}}{\mathrm{n}}=3 \mathrm{R} \end{aligned}$

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

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