The weight of body at earth's surface is W. At a depth half way to the centre of the earth, it will be (assuming uniform density in earth)

The correct option is W/2.

Let us understand the step-by-step physics behind how gravity and the weight of a body change as we go below the surface of the Earth.

Step 1: Understand the formula for acceleration due to gravity at a depth
The acceleration due to gravity on the surface of the Earth of radius $R$ is denoted by $g$.
When we go to a depth $d$ below the surface of the Earth, the acceleration due to gravity $g_d$ decreases because only the mass of the sphere of radius $(R-d)$ exerts a gravitational pull on the body (the outer shell of thickness $d$ exerts no net gravitational force at any interior point).
Assuming the Earth has a uniform density, the formula for acceleration due to gravity at depth $d$ is given by:

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

Step 2: Relate gravity to the weight of the body
The weight of a body of mass $m$ at the Earth's surface is:
$W=m·g$
Similarly, the weight of the body at depth $d$ is:
$W_d=m·g_d$
Substituting the expression for $g_d$ into the weight equation, we get:

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

Step 3: Calculate the weight at half the depth to the center
The question specifies that the body is at a depth halfway to the center of the Earth. This means:
$d=\frac{R}{2}$
Substituting this value of $d$ into our weight formula:

$W_d=W\left(1-\frac{\left(R/2\right)}{R}\right)$

Simplifying the expression inside the parentheses:

$W_d=W\left(1-\frac{1}{2}\right)$

$W_d=W·\frac{1}{2}=\frac{W}{2}$

Thus, the weight of the body at a depth halfway to the center of the Earth is $\frac{W}{2}$.