Keeping the mass of earth constant, if its radius is reduced to its of its (1/4) of its present 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 24h)

Question

Keeping the mass of earth constant, if its radius is reduced to its of its (1/4) of its present 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 24h) (A) 12 (B) 3 (C) 6 (D) 1.5

Tardigrade Admin · Accepted Answer

Assuming earth to be a solid sphere, its moment of inertia

I = \frac{2}{5} M R^{2}

As no external torque is acting on earth, thus its angular momentum is constant, i.e, Iw

=

constant

∴ \frac{2}{5} M R^{2} \times \frac{2 π}{T} =

constant

where

T

is the time period of earth's rotation

⟹ \frac{R _{1}^{2}}{T _{1}} = \frac{R _{2}^{2}}{T _{2}} (∵ M =

constant

)

Given :

R_{1} = R R_{2} = \frac{R}{4} T_{1} = 24 h rs

∴ \frac{R ^{2}}{24} = \frac{\frac{R ^{2}}{16}}{T _{2}} ⟹ T_{2} = \frac{24}{16} = 1.5 h rs

$12$

$3$

$6$

$1.5$

Q. Keeping the mass of earth constant, if its radius is reduced to its of its 41​ of its present 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 24h)

12

3

6

1.5

$12$

$3$

$6$

$1.5$