Question

Keeping the mass of the earth as constant, if its radius is reduced to 1/4 th of its initial value, then the period of revolution of the earth about its own axis and passing through the centre (in hours) is (assume the earth to be a solid sphere and its initial period of rotation as 24 h) [EAMCET 2014]

Easy

Solution

As we know that,
Time period of revolution of earth at initial condition,
$T = 2 π \sqrt{\frac{R_{e}}{g}}$ …(i)
So, if radius is reduced, $R^{'} = \frac{R_{e}}{4}$
Time period becomes $T^{'} = 2 π \sqrt{\frac{R_{e}}{4 g}}$ …(ii)
On dividing Eq. (i) by Eq. (ii), we get
$\frac{T}{T^{'}} = \frac{2 π \sqrt{R_{e} / g}}{2 π \sqrt{R_{e} / 4 g}} = 2$
$\Rightarrow T^{'} = \frac{T}{2} = \frac{24}{2} = 12 h$

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