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 determine how the period of rotation of the Earth changes when its radius decreases to one-quarter of its original value while keeping its mass constant. We will use the principle of conservation of angular momentum.

### Step-by-Step Solution:

1. **Identify Initial Parameters**:
- Let the original radius of the Earth be \( R \).
- The mass of the Earth is \( M \).
- The original period of rotation (one complete day) is \( T = 24 \) hours.

2. **Moment of Inertia Calculation**:
- The moment of inertia \( I \) of a solid sphere is given by:
\[ I = \frac{2}{5} M R^2 \]

3. **New Radius**:
- The new radius after shrinking is:
\[ R' = \frac{R}{4} \]

4. **New Moment of Inertia Calculation**:
- The new moment of inertia \( I' \) when the radius is \( R' \) is:
\[ I' = \frac{2}{5} M (R')^2 = \frac{2}{5} M \left(\frac{R}{4}\right)^2 = \frac{2}{5} M \frac{R^2}{16} = \frac{1}{80} M R^2 \]

5. **Angular Momentum Conservation**:
- Since no external torque is acting on the Earth, the angular momentum before and after shrinking must be equal:
\[ I \omega = I' \omega' \]
- Here, \( \omega \) is the original angular velocity and \( \omega' \) is the new angular velocity.
- The angular velocity is related to the period by:
\[ \omega = \frac{2\pi}{T} \quad \text{and} \quad \omega' = \frac{2\pi}{T'} \]

6. **Setting Up the Equation**:
- Substituting the values into the conservation of angular momentum equation:
\[ \left(\frac{2}{5} M R^2\right) \left(\frac{2\pi}{T}\right) = \left(\frac{1}{80} M R^2\right) \left(\frac{2\pi}{T'} \]
- Canceling \( M \) and \( R^2 \) from both sides:
\[ \frac{2}{5} \cdot \frac{2\pi}{T} = \frac{1}{80} \cdot \frac{2\pi}{T'} \]

7. **Simplifying the Equation**:
- Cancel \( 2\pi \):
\[ \frac{2}{5T} = \frac{1}{80T'} \]
- Cross-multiplying gives:
\[ 2 \cdot 80 T' = 5 T \]
- Thus:
\[ 160 T' = 5 T \]
- Rearranging for \( T' \):
\[ T' = \frac{5T}{160} = \frac{T}{32} \]

8. **Substituting the Original Period**:
- Since \( T = 24 \) hours:
\[ T' = \frac{24}{32} = \frac{3}{4} \text{ hours} = 1.5 \text{ hours} \]

### Final Answer:
The new period of rotation of the Earth, when its radius decreases to one-quarter of its original value, will be **1.5 hours**.