A small meteorite of mass 'm' travelling towards the centre of earth strikes the earth at the equator. The earth is a uniform sphere of mass 'M' and radius 'R'. The length of the day was 'T' before the meteorite struck. After the meteorite strikes the earth, the length of day increases (in sec) by

\frac{5 m T}{2 M}

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\frac{m}{M} T

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\frac{4 m T}{5 M}

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\frac{M}{3 m T}

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Solution

The correct option is A $\frac{5 m T}{2 M}$
$I_{1} ω_{1} = I_{2} ω_{2}$
$\frac{2}{5} M R^{2} \times (\frac{2 π}{T_{1}}) = (\frac{2}{5} M R^{2} + m R^{2}) \frac{2 π}{T_{2}}$
$\frac{T_{2}}{T_{1}} = 1 + \frac{5 m}{2 M}$
$\frac{T_{2} - T_{1}}{T_{1}} = \frac{5 m}{2 M}$
$T_{2} - T_{1} = \frac{5 m T}{2 M}$ $[i . e; T_{1} = T]$

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A small meteorite of mass 'm' travelling towards the centre of earth strikes the earth at the equator. The earth is a uniform sphere of mass 'M' and radius 'R'. The length of the day was 'T' before the meteorite struck. After the meteorite strikes the earth, the length of day increases (in sec) by

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A small meteorite of mass 'm' travelling towards the centre of earth strikes the earth at the equator. The earth is a uniform sphere of mass 'M' and radius 'R'. The length of the day was 'T' before the meteorite struck. After the meteorite strikes the earth, the length of day increases (in sec) by

The correct option is A 5mT2MI1ω1=I2ω2 25MR2×(2πT1)=(25MR2+mR2)2πT2 T2T1=1+5m2M T2−T1T1=5m2M T2−T1=5mT2M [i.e;T1=T]