If earth radius reduces to half of its present value then the time period of rotation will become:

Options

A. one fourth
B. one half
C. double
D. invariant

Answer

To solve the problem, we need to analyze the effect of reducing the Earth's radius to half on its rotational period. We will use the principle of conservation of angular momentum.

### Step-by-Step Solution:

1. **Understanding Angular Momentum**:
The angular momentum \( L \) of a rotating body is given by the product of its moment of inertia \( I \) and its angular velocity \( \omega \):
\[ L = I \cdot \omega \]

2. **Moment of Inertia of Earth**:
We can approximate the Earth as a solid sphere. The moment of inertia \( I \) for a solid sphere is given by:
\[ I = \frac{2}{5} m r^2 \]
where \( m \) is the mass of the Earth and \( r \) is its radius.

3. **Initial Angular Momentum**:
Let the initial radius of the Earth be \( r \) and the initial angular velocity be \( \omega_0 \). The initial angular momentum \( L_0 \) is:
\[ L_0 = I_0 \cdot \omega_0 = \frac{2}{5} m r^2 \cdot \omega_0 \]

4. **New Radius**:
If the radius is reduced to half, the new radius \( r' \) is:
\[ r' = \frac{r}{2} \]

5. **New Moment of Inertia**:
The new moment of inertia \( I' \) with the new radius is:
\[ I' = \frac{2}{5} m (r')^2 = \frac{2}{5} m \left(\frac{r}{2}\right)^2 = \frac{2}{5} m \cdot \frac{r^2}{4} = \frac{1}{10} m r^2 \]

6. **Conservation of Angular Momentum**:
Since no external torque acts on the Earth, angular momentum is conserved:
\[ L_0 = L' \quad \Rightarrow \quad \frac{2}{5} m r^2 \cdot \omega_0 = \frac{1}{10} m r^2 \cdot \omega' \]

7. **Solving for New Angular Velocity**:
Canceling \( m r^2 \) from both sides gives:
\[ \frac{2}{5} \omega_0 = \frac{1}{10} \omega' \quad \Rightarrow \quad \omega' = 4 \omega_0 \]

8. **Finding the New Time Period**:
The time period \( T \) of rotation is related to angular velocity by:
\[ T = \frac{2\pi}{\omega} \]
Thus, the new time period \( T' \) is:
\[ T' = \frac{2\pi}{\omega'} = \frac{2\pi}{4 \omega_0} = \frac{1}{4} \cdot \frac{2\pi}{\omega_0} = \frac{T_0}{4} \]
where \( T_0 \) is the original time period of the Earth.

9. **Conclusion**:
Therefore, if the Earth's radius reduces to half, the new time period of rotation will be:
\[ T' = \frac{T_0}{4} \]

### Final Answer:
The time period of rotation will become one-fourth of its present value.

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