At a certain depth "d " below surface of earth, value of acceleration due to gravity becomes four times that of its value at a height 3 R above earth surface. Where R is Radius of earth (Take R=6400 ~km ). The depth d is equal to

Gravitation PYQ — JEE Main 2023 · 31 Jan · Shift 1 · Q55 with Solution

View written solutionFree

Correct answer: D

For a depth $d$ inside Earth, assuming uniform density,

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

where $R$ is the radius of Earth.

If the height above the surface is $h = 3R$ , then distance from Earth's center is

r = R + 3R = 4R

So,

g_h = g\left(\frac{R}{4R}\right)^2 = \frac{g}{16}

The gravity at depth becomes four times the gravity at height $3R$ :

g_d = 4g_h

Substitute the expressions:

g\left(1 - \frac{d}{R}\right) = 4\cdot \frac{g}{16}

g\left(1 - \frac{d}{R}\right) = \frac{g}{4}

Cancel $g$ :

1 - \frac{d}{R} = \frac{1}{4}

\frac{d}{R} = 1 - \frac{1}{4} = \frac{3}{4}

d = \frac{3R}{4}

d = \frac{3}{4}\times 6400 = 4800\text{ km}

$4800\text{ km}$ corresponds to Option D.

Final Answer: $\boxed{4800\text{ km}}$