If the mass of earth and radius suddenly become 2 times and 14 th of the present value, the length of the day becomes

I1ω1=I2ω2⇒25M1R122πT1=25M2R222πT2⇒T2T1=M2M1×R2R12⇒T2=24×21×142=3 hr

If the mass of earth and radius suddenly become 2 times and 14 th of the present value, the length of the day becomes

I1ω1=I2ω2⇒25M1R122πT1=25M2R222πT2⇒T2T1=M2M1×R2R12⇒T2=24×21×142=3 hr

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If the mass of earth and radius suddenly become 2 times and $\frac{1}{4} t h$ of the present value, the length of the day becomes

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$24 h$

$6 h$

$\frac{3}{2} h$

$3 h$

answer is D.

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Detailed Solution

$I_{1} ω_{1} = I_{2} ω_{2}$

$\Rightarrow (\frac{2}{5} M_{1} R_{1}^{2}) (\frac{2 π}{T_{1}}) = (\frac{2}{5} M_{2} R_{2}^{2}) (\frac{2 π}{T_{2}})$

$\Rightarrow \frac{T_{2}}{T_{1}} = \frac{M_{2}}{M_{1}} \times {(\frac{R_{2}}{R_{1}})}^{2}$

$\Rightarrow T_{2} = 24 \times \frac{2}{1} \times {(\frac{1}{4})}^{2} = 3 h r$

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If the mass of earth and radius suddenly become 2 times and 14 th of the present value, the length of the day becomes

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If the mass of earth and radius suddenly become 2 times and $\frac{1}{4} t h$ of the present value, the length of the day becomes