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

<p>To solve this question, we first need to understand how gravitational force (and hence weight) changes with depth under the surface of the Earth. The gravitational force at a depth $d$ is given by the formula:</p> <p>$F = F_0 \left(1 - \frac{d}{R}\right)$</p> <p>where $F_0$ is the gravitational force (or the weight) at the surface, $R$ is the radius of the Earth, and $d$ is the depth below the Earth's surface.</p> <p>In your question, the body weighs 300 N on the surface, so $F_0 = 300$ N. It is taken to a depth of $\frac{R}{4}$ under the surface. Therefore, $d = \frac{R}{4}$.</p> <p>Substituting these values into our formula, we get:</p> <p>$F = 300 \left(1 - \frac{\frac{R}{4}}{R}\right)$</p> <p>$F = 300 \left(1 - \frac{1}{4}\right)$</p> <p>$F = 300 \left(\frac{3}{4}\right)$</p> <p>$F = 225 \, \text{N}$</p> <p>Therefore, at a depth of $\frac{R}{4}$ under the surface of the Earth, the body would weigh 225 N. The correct option is <strong>Option D</strong>.</p>