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:

<p>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.</p> <p>The weight of an object at a depth $d$ from the Earth&#39;s surface is given by:</p> <p>$W_d = W_e (1 - \frac{d}{R})$,</p> <p>where:</p> <ul> <li>$W_d$ is the weight at depth $d$,</li> <li>$W_e$ is the weight at the Earth&#39;s surface (i.e., the weight of the object), and</li> <li>$R$ is the radius of the Earth.</li> </ul> <p>In this case, we&#39;re given that $W_e = 400 \, \text{N}$, and we&#39;re asked to find the weight at a depth of $d = \frac{R}{2}$. Substituting these values into the formula, we get:</p> <p>$W_d = 400 \, \text{N} (1 - \frac{1}{2}) = 200 \, \text{N}$.</p> <p>Therefore, the weight of the body when taken to a depth half of the radius of the Earth is 200 N.</p>