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Keeping the mass of earth as constant, if its radius is reduced to 1/4^th of its initial value, then the period of revolution of earth about its own axis and passing through the centre, in hours, is (Assure earth to be a solid sphere and its initial period of rotation as 24 hrs.

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1.5

answer is A.

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Detailed Solution

M = constant, $R_{2} = \frac{R_{1}}{4}$

T₁= 24hr, T₂ = ?

$I_{1} ω_{1} = I_{2} ω_{2}$

$\frac{2}{5} {MR}_{1}^{2} \times \frac{2 π}{T_{1}} = \frac{2}{5} M {R_{2}}^{2} \times \frac{2 π}{T_{2}}$

$\frac{R_{1}}{24} = {(\frac{R_{1}}{4})}^{2} \times \frac{1}{T_{2}}$

$\Rightarrow T_{2} = \frac{24}{16} = 1.5 hr$

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