Q 11-07-040JEE MainJEE Main 2026 (5 Apr, Shift 1)Medium
When one moves from a point $16$ km below the earth's surface to a point $16$ km above the earth's surface. The change in g is approximately $\alpha\%$. The value of $\alpha$ is ______.
(Take radius of the earth $= 6400$ km.)
Answer: (B) $0.25$
Concept. For small heights and depths: going up $h$ reduces $g$ by the fraction $\dfrac{2h}{R}$; going down $d$ reduces it by $\dfrac dR$. So $g$ is largest at the surface and falls on both sides, faster going up.
Step 1. At $16$ km depth. $g_1=g\left(1-\dfrac{16}{6400}\right)=g(1-0.0025)$.
Step 2. At $16$ km height. $g_2=g\left(1-\dfrac{2\times16}{6400}\right)=g(1-0.005)$.
Step 3. Change. $g_1-g_2=0.0025\,g$, which is $0.25\%$ of $g$.
Answer. $\alpha=0.25$ (Option B).
Solution by Sreeraj P, M.Sc Physics
