Q.39. The depth at which acceleration due to gravity becomes \frac{g}{2} is (R= radius of earth, g= acceleration due to gravity on earth’s surface, n is integer)
A. \frac{R(1-2n)}{n}
B. \frac{R(1-n)}{2n}
C. \frac{R(n-1)}{n}
D. \frac{R(2n-1)}{2n}
Answer :- D. \frac{R(2n-1)}{2n}
Explanation :-
The gravitational acceleration at depth is given as g_d=g\left[1-\frac{d}{R}\right]
Given \mathrm{g}_{\mathrm{d}}=\frac{\mathrm{g}}{2 \mathrm{n}}
\therefore \quad \frac{g}{2 n}=g\left[1-\frac{d}{R}\right] \frac{d}{R}=1-\frac{1}{2 n} d=\left[\frac{2 n-1}{2 n}\right] R