Keeping the mass of the earth as constant, if its radius is reduced to 1/4th of its initial value, then the period of revolution of the earth about its own axis and passing through the centre, (in hours) is (assume the earth to be a solid sphere and its initial period of rotation as 24 h)

Options

A. 12
B. 3
C. 6
D. 1.5

Explanation

To solve the problem, we need to find the new period of revolution (T2) of the Earth about its own axis when its radius is reduced to 1/4th of its initial value while keeping its mass constant. We will use the relationship between the period of rotation and the radius of the Earth.

### Step-by-Step Solution:

1. **Understand the Relationship**:
The period of rotation (T) of a solid sphere is related to its radius (R) by the formula:
\[ T \propto R^{3/2} \]
This means that if the radius changes, the period will change according to this proportionality.

2. **Define Initial Conditions**:
Let:
- \( T_1 = 24 \) hours (initial period of rotation)
- \( R_1 = R \) (initial radius of the Earth)

3. **Define New Conditions**:
The new radius \( R_2 \) is given as:
\[ R_2 = \frac{R_1}{4} = \frac{R}{4} \]

4. **Apply the Proportionality**:
Using the relationship:
\[ \frac{T_1}{T_2} = \left(\frac{R_1}{R_2}\right)^{3/2} \]
We can substitute \( R_2 \):
\[ \frac{T_1}{T_2} = \left(\frac{R}{\frac{R}{4}}\right)^{3/2} = \left(4\right)^{3/2} \]

5. **Calculate \( 4^{3/2} \)**:
\[ 4^{3/2} = (2^2)^{3/2} = 2^{3} = 8 \]
Therefore:
\[ \frac{T_1}{T_2} = 8 \]

6. **Solve for \( T_2 \)**:
Rearranging gives:
\[ T_2 = \frac{T_1}{8} = \frac{24 \text{ hours}}{8} = 3 \text{ hours} \]

### Final Answer:
The new period of revolution of the Earth about its own axis, when its radius is reduced to 1/4th of its initial value, is **3 hours**.

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