MHT CET · Physics · Rotational Motion

A body is situated on the surface of the earth becomes weightless at equator when the rotational kinetic energy of the earth reaches a critical value ' \(K\) '. The value of \(K\) is given by [ \(\mathrm{g}\) = gravitational acceleration on earth's surface, \(M=\) mass of the earth and \(\mathrm{R}=\) radius of the earth]

A \(\frac{1}{2} \mathrm{MgR}\)
B \(\frac{1}{3} \mathrm{MgR}\)
C \(\frac{1}{4} \mathrm{MgR}\)
D \(\frac{1}{5} \mathrm{MgR}\)

Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{1}{5} \mathrm{MgR}\)

Step-by-step Solution

Detailed explanation

A body will become weightless at equator if \(R \omega^2=g\) or \(\omega^2=\frac{g}{R}\) Kinetic energy of the earth, \(\mathrm{K}=\frac{1}{2} \mathrm{I} \omega^2\)
For a solid sphere, \(\mathrm{I}=\frac{2}{5} \mathrm{MR}^2\)
(M is mass of the earth)
\(\therefore \mathrm{K}=\frac{1}{2} \times \frac{2}{5} \mathrm{MR}^2 \times \frac{\mathrm{g}}{\mathrm{R}}=\frac{1}{5} \mathrm{MgR}\)

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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