JEE Main 2026 — Gravitation Question with Solution

The acceleration due to gravity at a depth $d$ below the earth's surface is given by:

$g_d=g\left(1-\frac{d}{R}\right)$

The acceleration due to gravity at a height $h$ above the earth's surface (for $h\ll R$) is given by:

$g_h=g\left(1-\frac{2h}{R}\right)$

Given $d=16$ km and $h=16$ km, we can find the values of $g_d$ and $g_h$:

$g_d=g\left(1-\frac{16}{6400}\right)=g\left(1-\frac{1}{400}\right)$

$g_h=g\left(1-\frac{2\times 16}{6400}\right)=g\left(1-\frac{2}{400}\right)$

The magnitude of the change in acceleration due to gravity when moving from the depth $d$ to the height $h$ is:

$\Delta g=g_d-g_h=g\left(1-\frac{1}{400}\right)-g\left(1-\frac{2}{400}\right)$

$\Delta g=g\left(\frac{2}{400}-\frac{1}{400}\right)=\frac{g}{400}$

The percentage change in $g$ is approximately:

$\frac{\Delta g}{g}\times 100=\frac{\frac{g}{400}}{g}\times 100=\frac{100}{400}=0.25\%$

Thus, the value of $\alpha$ is $0.25$.

Answer: $0.25$