# If the Earth is compressed to half of its radius and its mass remains the same, then the day will be of

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If the Earth is compressed to half of its radius and its mass remains the same, then the day will be of
1. 6 Hours
2. 12 Hours
3. 24 Hours
4. 36 Hours

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Correct Answer - Option 1 : 6 Hours

Concept:

Conservation of angular momentum:

A rigid spinning object continues to spin at a constant rate and with a fixed orientation unless influenced by the application of external torque.

i.e., $\tau = \frac{{dL}}{{dt}}$ and if momentum is conserved there would be no external torque acting on it.

$\tau = 0 = \frac{{dL}}{{dt}}\;or\;\frac{{{\rm{\Delta }}L}}{{{\rm{\Delta }}t}}$

• This means that the change in angular momentum is zero if there is no external torque i.e., ΔL = 0. Thus, for a closed system, angular momentum is constant

The angular momentum is given as

L = I ω

I is a moment of inertia, ω is angular velocity.

Calculation:

The moment of inertia of Earth is

$I = \frac{2}{5} MR^2$

M is mass, R is the radius

Angular speed is given as

$ω = \frac{2 \pi }{T}$

In process of the sinking of the earth, no external torque is applied.

Mass is same, radius changed

So,

I1 ω 1 = I2 ω 2

$\implies \frac{2}{5} MR^2 (\frac{2 \pi }{T}) = \frac{2}{5} MR'^2 (\frac{2 \pi }{T'})$

$\implies (\frac{R^2 }{T}) = (\frac{R'^2 }{T'})$

The new radius R' = R / 2

New Time Period T' , which we have to find

So,

$\implies (\frac{R^2 }{T}) = (\frac{(R'/2)^2 }{T'})$

$\implies T' = \frac{T}{4}$

T is 24 hr in normal conditon.

So, T' = 24 hr / 4 = 6 hr

So, the correct option is 6 hr.