MHT CET · Physics · Gravitation
A mine is located at depth \(\frac{\mathrm{R}}{3}\) below earth's surface. The acceleration due to gravity at that depth in mine is ( \(\mathrm{R}=\) radius of earth, \(\mathrm{g}\) = acceleration due to gravity)
- A \(\mathrm{g}\)
- B \(3 g\)
- C \(\frac{2 \mathrm{~g}}{3}\)
- D \(\frac{\mathrm{g}}{3}\)
Answer & Solution
Correct Answer
(C) \(\frac{2 \mathrm{~g}}{3}\)
Step-by-step Solution
Detailed explanation
The acceleration due to gravity at depth \(\mathrm{d}\) is \(\mathrm{g}_{\mathrm{d}}=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)\)
given \(\mathrm{d}=\mathrm{R} / 3\)
\(\therefore \quad \mathrm{g}_{\mathrm{d}}=\mathrm{g}\left(1-\frac{(\mathrm{R} / 3)}{\mathrm{R}}\right)=\frac{2}{3} \mathrm{~g}\)
given \(\mathrm{d}=\mathrm{R} / 3\)
\(\therefore \quad \mathrm{g}_{\mathrm{d}}=\mathrm{g}\left(1-\frac{(\mathrm{R} / 3)}{\mathrm{R}}\right)=\frac{2}{3} \mathrm{~g}\)
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