Question

Assuming the earth to be a sphere of uniform mass density, a body weighed 300 N on the surface of earth. How much it would weigh at R/4 depth under surface of earth ?

• A. 75 N
• B. 375 N
• C. 300 N
• D. 225 N

Answer 1

The Correct Option is D

At the surface:
\[mg = 300 \, \text{N}\]
\[m = \frac{300}{g_s}\]
At depth \( \frac{R}{4} \):
\[g_d = g_s \left( 1 - \frac{d}{R} \right)\]
where \( d = \frac{R}{4} \).
\[g_d = g_s \left( 1 - \frac{R}{4R} \right) = g_s \cdot \frac{3}{4}\]
The weight at depth \( \frac{R}{4} \) is:
\[\text{Weight} = m \times g_d = m \times \frac{3 g_s}{4}\]
\[= \frac{3}{4} \times 300 = 225 \, \text{N}\]

Answer 2

To find the weight of a body at a depth \( \frac{R}{4} \) beneath the Earth's surface, we must understand how gravitational force changes with depth inside the Earth.

Assuming the Earth is a sphere of uniform density, the gravitational force inside Earth at a depth \( d \) is given by the formula:

\(W_d = W_0 \left(1 - \frac{d}{R}\right)\)

Where:

• \(W_d\) is the weight at depth \( d \).
• \(W_0\) is the weight at the Earth's surface.
• \(d\) is the depth below the surface, \( \frac{R}{4} \) in this case.
• \(R\) is the radius of the Earth.

Given:

• \(W_0 = 300 \, \text{N}\)
• \(d = \frac{R}{4}\)

Substitute into the formula:

\(W_d = 300 \left(1 - \frac{1}{4}\right) = 300 \left(\frac{3}{4}\right) = 225 \, \text{N}\)

Thus, the weight of the body at a depth of \( \frac{R}{4} \) under the surface of the Earth would be 225 N.

Therefore, the correct answer is 225 N.