If the radius of earth shrinks to kR (k lt 1), where R is the radius of Earth, then the time period of rotation of Earth, about its axis will become

Options

A. `(24)/(k)` hrs
B. 24 k hrs
C. `24k^(2)` hrs
D. `(24)/(k^(2))` hrs

Answer

To solve the problem, we need to determine how the time period of Earth's rotation changes when its radius shrinks to \( kR \), where \( k < 1 \).

### Step-by-Step Solution:

1. **Understanding Angular Momentum**:
Since no external torque is applied during the shrinking of the Earth, the angular momentum \( L \) remains constant. The angular momentum \( L \) is given by:
\[ L = I \omega \]
where \( I \) is the moment of inertia and \( \omega \) is the angular speed.

2. **Moment of Inertia of a Solid Sphere**:
For a solid sphere, the moment of inertia \( I \) is given by:
\[ I = \frac{2}{5} m r^2 \]
where \( m \) is the mass of the Earth and \( r \) is its radius.

3. **Angular Speed Relation**:
The angular speed \( \omega \) is related to the time period \( T \) by:
\[ \omega = \frac{2\pi}{T} \]

4. **Setting Up the Equation**:
Since \( L \) is constant, we can write:
\[ I_{\text{initial}} \omega_{\text{initial}} = I_{\text{final}} \omega_{\text{final}} \]
Substituting the expressions for \( I \) and \( \omega \):
\[ \frac{2}{5} m R^2 \cdot \frac{2\pi}{T_{\text{initial}}} = \frac{2}{5} m (kR)^2 \cdot \frac{2\pi}{T_{\text{final}}} \]

5. **Simplifying the Equation**:
The mass \( m \) and the factor \( \frac{2}{5} \) cancel out from both sides:
\[ R^2 \cdot \frac{1}{T_{\text{initial}}} = (kR)^2 \cdot \frac{1}{T_{\text{final}}} \]
This simplifies to:
\[ \frac{R^2}{T_{\text{initial}}} = k^2 \frac{R^2}{T_{\text{final}}} \]

6. **Cancelling \( R^2 \)**:
Since \( R^2 \) is common on both sides and \( R \neq 0 \), we can cancel it:
\[ \frac{1}{T_{\text{initial}}} = k^2 \frac{1}{T_{\text{final}}} \]

7. **Rearranging for \( T_{\text{final}} \)**:
Rearranging gives:
\[ T_{\text{final}} = k^2 T_{\text{initial}} \]

8. **Substituting the Initial Time Period**:
The initial time period \( T_{\text{initial}} \) is given as 24 hours:
\[ T_{\text{final}} = k^2 \cdot 24 \text{ hours} \]

### Final Answer:
Thus, the time period of rotation of Earth about its axis when its radius shrinks to \( kR \) will be:
\[ T_{\text{final}} = 24k^2 \text{ hours} \]