If the Earth shrinks such that its density becomes 8 times its present value then the new duration of the day in hours will be : (assume that there is no change in axis of rotation of the earth)

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Solution

The correct option is C 6
If M is mass of the earth, then clearly $M_{i} = M_{f}$ .
(where $M_{i}$ and $M_{f}$ represent the initial and final mass of the earth)
Now $d = \frac{M}{V}$ , where V will be the volume of a sphere, given by $V = \frac{4}{3} π R^{3}$ , where R is the radius of earth.
Since Mass is constant , $d_{i} R_{i}^{3} = d_{f} R_{f}^{3}$ .
putting $d_{f}$ = $8 d_{i}$ , we get: $R_{f} = \frac{R_{i}}{2}$
Now since there is no external torque applied, by conservation of angular momentum, we get:
$L_{i} = L_{f}$
hence, $I_{i} ω_{i} = I_{f} ω_{f}$
hence, $\frac{2}{5} M_{i} R_{i}^{2} \frac{2 π}{T_{i}} = \frac{2}{5} M_{f} R_{f}^{2} \frac{2 π}{T_{f}}$
putting $M_{i} = M_{f}$ , $T_{i} = 24 h$ , $R_{f} = \frac{R_{i}}{2}$ ,
we get $T_{f} = 6 h$

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