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
- A 2 R
- B 3 R
- C 4 R
- D 5 R
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,
\(\begin{aligned}
\quad \frac{\mathrm{g}_{\mathrm{h}}}{\mathrm{~g}} & =\left(\frac{\mathrm{n}}{\mathrm{n}+1}\right)^2=\frac{1}{16} \Rightarrow \frac{\mathrm{n}}{\mathrm{n}+1}=\frac{1}{4} \\
\therefore \quad \mathrm{n} & =\frac{1}{3} \\
\mathrm{~h} & =\frac{\mathrm{R}}{\mathrm{n}}=3 \mathrm{R}
\end{aligned}\)
\(\begin{aligned}
\quad \frac{\mathrm{g}_{\mathrm{h}}}{\mathrm{~g}} & =\left(\frac{\mathrm{n}}{\mathrm{n}+1}\right)^2=\frac{1}{16} \Rightarrow \frac{\mathrm{n}}{\mathrm{n}+1}=\frac{1}{4} \\
\therefore \quad \mathrm{n} & =\frac{1}{3} \\
\mathrm{~h} & =\frac{\mathrm{R}}{\mathrm{n}}=3 \mathrm{R}
\end{aligned}\)
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