JEE Main 2023 — Gravitation Question with Solution

The gravitational field inside a uniform spherical body varies linearly with distance from the center. If we consider the Earth to be such a body, then the gravitational field strength (and hence weight) of an object would decrease linearly as we go deeper inside the Earth.

The weight of an object at a depth $d$ from the Earth's surface is given by:

$W_d=W_e(1-\frac{d}{R})$,

where:

• $W_d$ is the weight at depth $d$,
• $W_e$ is the weight at the Earth's surface (i.e., the weight of the object), and
• $R$ is the radius of the Earth.

In this case, we're given that $W_e=400\text{N}$, and we're asked to find the weight at a depth of $d=\frac{R}{2}$. Substituting these values into the formula, we get:

$W_d=400\text{N}(1-\frac{1}{2})=200\text{N}$.

Therefore, the weight of the body when taken to a depth half of the radius of the Earth is 200 N.