MHT CET · Physics · Gravitation
A body of mass ' \(\mathrm{m}\) ' is raised through a height above the earth's surface so that the increase in potential energy is \(\frac{\mathrm{mgR}}{5}\). The height to which the body is raised is ( \(\mathrm{R}=\) radius of earth, \(\mathrm{g}=\) acceleration due to gravity)
- A \(\mathrm{R}\)
- B \(\frac{\mathrm{R}}{2}\)
- C \(\frac{\mathrm{R}}{4}\)
- D \(\frac{\mathrm{R}}{8}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{R}}{4}\)
Step-by-step Solution
Detailed explanation
When a particle of mass \(m\) is taken from the Earth's surface to a height \(h=n R\), then the change in P.E. can be calculated as,
\(\Delta \mathrm{U}=\operatorname{mgR}\left(\frac{\mathrm{n}}{\mathrm{n}+1}\right) \)
\( \therefore \frac{\mathrm{mgR}}{5}=\mathrm{mgR}\left(\frac{\mathrm{n}}{\mathrm{n}+1}\right) \)
\( \therefore \mathrm{n}+1=5 \mathrm{n} \mathrm{n}=\frac{1}{4} \)
\( \therefore \mathrm{h}=\frac{\mathrm{R}}{4}\)
\(\Delta \mathrm{U}=\operatorname{mgR}\left(\frac{\mathrm{n}}{\mathrm{n}+1}\right) \)
\( \therefore \frac{\mathrm{mgR}}{5}=\mathrm{mgR}\left(\frac{\mathrm{n}}{\mathrm{n}+1}\right) \)
\( \therefore \mathrm{n}+1=5 \mathrm{n} \mathrm{n}=\frac{1}{4} \)
\( \therefore \mathrm{h}=\frac{\mathrm{R}}{4}\)
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