Question

A body weighs 200 N on the surface of the earth. How much will it weigh half way down to the centre of the earth? [NEET 2019]

Easy

Solution

Given, weight of the body, $ω$ = 200 N
As we know $ω$ = mg, where m is the mass of the body and B (= 10 ms^-2) is acceleration due to gravity of the body at the surface of the earth.
Since, mass m remains constant irrespective of the position of the body on the earth. However, B is not constant and its value at a depth d below the earth's surface is given as
$g = g (1 - \frac{d}{R})$ ... (i)
where, R is the radius of the earth.
Multiplying m on both sides of Eq. (i), we get
$m g^{'} = m g (1 - \frac{d}{R})$
Now, the weight of the body at half-way down $(i.e. d = \frac{R}{2})$ to the centre of the earth is
$m g^{'} = 200 \times (1 - \frac{R / 2}{R})$
$= 200 (1 - \frac{1}{2}) = 200 \times \frac{1}{2} = 100 N$
$∴$ The body will weigh 100 N half-way down to the centre of the earth.

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