MHT CET · Physics · Gravitation

A hole is drilled half way to the centre of the earth. A body weighs 300 N on the surface of the earth. How much will, it weigh at the bottom of the hole?

A 200 N
B 250 N
C 120 N
D 150 N

Verified Solution

Answer & Solution

Correct Answer

(D) 150 N

Step-by-step Solution

Detailed explanation

Given, distance of bottom of hole from the surface of earth (d) = half the radius of earth =

\frac{R_{e}}{2}

If g be the value of gravitational acceleration on the surface of earth, then weight of body

m g = 300 N

If g’ be the gravitational acceleration at the bottom of hole, then

g^{'} = g (1 - \frac{d}{R_{e}}) = g (1 - \frac{\frac{R_{e}}{2}}{R_{e}}) \Rightarrow g^{'} = \frac{g}{2}

∴

Weight of the body on the bottom of hole,

m g^{'} = \frac{m g}{2} = \frac{300}{2} = 150 N

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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