Question

Assuming the earth to be a sphere of uniform mass density, the weight of a body at a depth $d = \frac{R}{2}$ from the surface of earth, if its weight on the surface of earth is 200 N, will be : (Given R = Radius of earth)

• A. 400 N
• B. 500 N
• C. 300 N
• D. 100 N

Answer 1

Answer: (D)

100 N

Answer 2

The acceleration due to gravity at a depth $d$ from the surface of the Earth is given by the formula:

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

where $g$ is the acceleration due to gravity on the surface of the Earth and $R$ is the radius of the Earth.

The weight of a body is given by $W = mg$.

Let $W_s$ be the weight of the body on the surface and $W_d$ be the weight at depth $d$.

$W_s = mg$ $W_d = mg_d$

Substituting the expression for $g_d$:

$W_d = m \cdot g \left(1 - \frac{d}{R}\right) = (mg) \left(1 - \frac{d}{R}\right)$

Since $W_s = mg$, we have:

$W_d = W_s \left(1 - \frac{d}{R}\right)$

Given in the question:

Weight on the surface, $W_s = 200$ N Depth from the surface, $d = \frac{R}{2}$

Substitute these values into the formula for $W_d$:

$W_d = 200 \text{ N} \times \left(1 - \frac{R/2}{R}\right)$ $W_d = 200 \text{ N} \times \left(1 - \frac{1}{2}\right)$ $W_d = 200 \text{ N} \times \left(\frac{1}{2}\right)$ $W_d = 100 \text{ N}$

The weight of the body at a depth $d = R/2$ from the surface of the Earth is 100 N.