Question

If the earth suddenly shrinks to \(\frac{1}{ 64}\) th of its original volume with its mass remaining the same, the period of rotation of earth becomes \(\frac{24}{ x}\) h. The value of x is .

Hint:

Remember the conservation of angular momentum: Iω = constant. Changes in the moment of inertia (I) directly affect the angular velocity (ω) and thus the period of rotation.

Answer 1

Step 1: Simplify the Equation 

The angular momentum conservation equation is given by:

\[ \frac{2}{5} M R^2 \omega_1^2 = \frac{2}{5} M \left(\frac{R}{4}\right)^2 \omega_2^2 \]

Cancel the mass \( M \) and constant \( \frac{2}{5} \):

\[ \frac{\omega_1}{\omega_2} = \left(\frac{R}{R/4}\right)^2 = \frac{1}{16} \]

Step 2: Relate Angular Velocity to Time Period

The relationship between angular velocity and time period is:

\[ \frac{\omega_1}{\omega_2} = \frac{T_2}{T_1} \]

Substituting \( \frac{\omega_1}{\omega_2} = \frac{1}{16} \) and \( T_1 = 24 \):

\[ \frac{1}{16} = \frac{T_2}{24} \]

Step 3: Solve for \( T_2 \)

Rearrange to find \( T_2 \):

\[ T_2 = \frac{24}{16} \]

\[ T_2 = x = 16 \]

Final Answer:

The value of \( x \) is 16.