JEE PYQ: Gravitation Question 25
Question 25 - 2020 (05 Sep Shift 1)
The value of acceleration due to gravity is $g_1$ at a height $h = \frac{R}{2}$ ($R$ = radius of the earth) from the surface of the earth. It is again equal to $g_1$ at a depth $d$ below the surface of the earth. The ratio $\left(\frac{d}{R}\right)$ equals:
(1) $\frac{4}{9}$
(2) $\frac{5}{9}$
(3) $\frac{1}{3}$
(4) $\frac{7}{9}$
Show Answer
Answer: (2)
Solution
At height $h = R/2$: $g_1 = \frac{g}{(3/2)^2} = \frac{4g}{9}$. At depth $d$: $g_1 = g(1-d/R)$. So $\frac{4}{9} = 1 - d/R$, giving $d/R = 5/9$.
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