Question

If the radius of earth contracts 1/n of its present day value, the length of the day will be approximately

Answer 1

Answer:

Given:

Radius of Earth becomes 1/n times as the original value

To find:

New length of day

Concept:

We know that earth is rotating around its geometric axis. Since there is involvement of no external torque , we can say that the Angular Momentum of Earth remains constant.

$\sf{ \blue{L = constant}}$

$\sf{ \blue{ = > I \times \omega = constant}}$

Calculation:

Considering Earth as a solid sphere , we can say that initial Moment of Inertia is as follows :

$\green{I = \dfrac{2}{5} m {r}^{2} }$

New radius is r/n

$\green{I2 = \dfrac{2}{5} m \bigg({ \dfrac{r}{n} \bigg) }^{2} }$

$\green{ = > I2 = \dfrac{I}{ {n}^{2} } }$

Now putting this value in the Angular Momentum equation :

$\sf{ \blue{ = > I \times \omega =I2 \times \omega2 }}$

$\sf{ \blue{ = > I \times \omega = \dfrac{I}{ {n}^{2} } \times \omega2 }}$

$\sf{ \blue{ = > \omega2 = {n}^{2} \times \omega}}$

Time period is given as follows :

$\boxed{ \orange{T = \dfrac{2\pi}{ \omega} = 24 \: hr}}$

New time period will be :

$\boxed{ \orange{T2 = \dfrac{2\pi}{ {n}^{2} \omega} = \dfrac{24}{( {n}^{2} )} \: hr}}$

So final answer is :

$\boxed{ \red{ \huge{ \sf{ \bold{T2 = \dfrac{24}{ ({n}^{2} )}}}}}}$

Answer 2

Answer:

24/n² hrs

Explanation:

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