The depth $d$ at which the acceleration due to gravity becomes $\frac{g}{n}$ is (where $R$ is the radius of the Earth,$g$ is the acceleration due to gravity at the surface,and $n$ is an integer).
- A$\frac{R(n-1)}{n}$
- B$\frac{R(n+1)}{n}$
- C$\frac{R(n-1)^2}{n}$
- D$\frac{R(n+1)^2}{n}$
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What is the depth at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the surface of the Earth? (radius of Earth $= R$)
MediumNEET 2020
View SolutionThe acceleration due to gravity near the surface of a planet of radius $R$ and density $d$ is proportional to
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