JEE MAIN - Physics (2023 - 8th April Morning Shift - No. 16)

The weight of a body on the earth is $$400 \mathrm{~N}$$. Then weight of the body when taken to a depth half of the radius of the earth will be:
300 N
200 N
100 N
Zero

Explanation

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.

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