If the earth shrinks such that its mass does not change but radius decreases to one-quarter of its original value then one complete day will take

Options

A. 96 h
B. 48 h
C. 6 h
D. 1.5 h

Answer

To solve the problem, we need to analyze the situation where the Earth's radius decreases to one-quarter of its original value while its mass remains unchanged. We will use the principles of conservation of angular momentum and the formulas for moment of inertia and angular velocity.

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- Let the original radius of the Earth be \( R \).
- The mass of the Earth is \( M \).
- The initial moment of inertia \( I_i \) of a solid sphere is given by:
\[ I_i = \frac{2}{5} M R^2 \]
- The initial angular velocity \( \omega_i \) can be calculated using the period of rotation (1 day = 24 hours):
\[ \omega_i = \frac{2\pi}{T_i} = \frac{2\pi}{24 \times 3600} \text{ rad/s} \]

2. **Determine the Final Conditions**:
- The new radius \( R_f \) is:
\[ R_f = \frac{R}{4} \]
- The final moment of inertia \( I_f \) is:
\[ I_f = \frac{2}{5} M R_f^2 = \frac{2}{5} M \left(\frac{R}{4}\right)^2 = \frac{2}{5} M \frac{R^2}{16} = \frac{1}{80} M R^2 \]

3. **Apply Conservation of Angular Momentum**:
- Since there is no external torque acting on the Earth, angular momentum is conserved:
\[ I_i \omega_i = I_f \omega_f \]
- Substituting the expressions for moment of inertia:
\[ \frac{2}{5} M R^2 \cdot \omega_i = \frac{1}{80} M R^2 \cdot \omega_f \]
- Cancel \( M \) and \( R^2 \) from both sides:
\[ \frac{2}{5} \omega_i = \frac{1}{80} \omega_f \]

4. **Solve for Final Angular Velocity**:
- Rearranging gives:
\[ \omega_f = \frac{2}{5} \cdot 80 \cdot \omega_i = 32 \omega_i \]

5. **Relate Angular Velocity to Time Period**:
- The relationship between angular velocity and the time period is:
\[ \omega = \frac{2\pi}{T} \]
- Therefore, the final period \( T_f \) is:
\[ T_f = \frac{2\pi}{\omega_f} = \frac{2\pi}{32 \omega_i} \]
- Since \( T_i = 24 \text{ hours} \):
\[ T_f = \frac{T_i}{32} = \frac{24 \text{ hours}}{32} = \frac{3}{4} \text{ hours} = 1.5 \text{ hours} \]

### Final Answer:
One complete day will take **1.5 hours**.