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MHT CET · Physics · Gravitation

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

  1. A 2 R
  2. B 3 R
  3. C 4 R
  4. D 5 R
Verified Solution

Answer & Solution

Correct Answer

(B) 3 R

Step-by-step Solution

Detailed explanation

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