Gravitation

Question: The depth d at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the surface, is [R = radius of the earth]

1) $\frac{R}{n}$

2) $R\,\left( {\frac{{n - 1}}{n}} \right)$

3) $\frac{R}{{{n^2}}}$

4) $R\,\left( {\frac{n}{{n + 1}}} \right)$

Solution:

Explanation: No Explanation

Acceleration Due to Gravity

Rate this question:

★ ★ ★ ★ ★

Average rating: (0 votes)