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Q.

What should be the angular velocity of rotation of Earth about its own axis so that the weight of a body at the equator reduces to $\frac{3}{5}$ of its present value? (Take R as the radius of the Earth)

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a

$\sqrt{\frac{g}{3 R}}$

b

$\sqrt{\frac{2 g}{3 R}}$

c

$\sqrt{\frac{2 g}{5 R}}$

d

$\sqrt{\frac{2 g}{7 R}}$

answer is C.

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Detailed Solution

${mg}^{'} = mg - {mRω}^{2} \cos^{2} φ$

Now, $\frac{3}{5} mg = mg - {mRω}^{2}$

or ${mRω}^{2} = mg - \frac{3}{5} mg$

or ${mRω}^{2} = \frac{2}{5} mg or ω = \sqrt{\frac{2 g}{5 R}}$

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What should be the angular velocity of rotation of Earth about its own axis so that the weight of a body at the equator reduces to 35 of its present value? (Take R as the radius of the Earth)