If the radius of the earth becomes half of its present value (mass remaining the same ), the new length of the day would be

Options

A. 6 hours
B. 12 hours
C. 48 hours
D. 96 hours

Answer

To solve the problem, we need to determine the new length of the day if the radius of the Earth becomes half of its present value while keeping the mass constant. We will use the principles of angular momentum and the relationship between angular velocity and the time period.

### Step-by-Step Solution:

1. **Define Variables**:
- Let \( R \) be the original radius of the Earth.
- Let \( M \) be the mass of the Earth.
- Let \( \omega \) be the original angular velocity of the Earth.
- Let \( T \) be the original time period (length of the day), which is 24 hours.

2. **Moment of Inertia**:
The moment of inertia \( I \) of a solid sphere is given by:
\[ I = \frac{2}{5} M R^2 \]
For the new Earth with radius \( \frac{R}{2} \), the moment of inertia \( I' \) becomes:
\[ I' = \frac{2}{5} M \left(\frac{R}{2}\right)^2 = \frac{2}{5} M \frac{R^2}{4} = \frac{1}{10} M R^2 \]

3. **Angular Momentum Conservation**:
Since there is no external torque acting on the Earth, the angular momentum before and after the change must be equal:
\[ I \omega = I' \omega' \]
Substituting the values of \( I \) and \( I' \):
\[ \frac{2}{5} M R^2 \omega = \frac{1}{10} M R^2 \omega' \]
We can cancel \( M R^2 \) from both sides (since \( M \) and \( R \) are not zero):
\[ \frac{2}{5} \omega = \frac{1}{10} \omega' \]

4. **Solve for New Angular Velocity**:
Rearranging the equation gives:
\[ \omega' = 4 \omega \]
This means the new angular velocity \( \omega' \) is four times the original angular velocity \( \omega \).

5. **Relate Angular Velocity to Time Period**:
The angular velocity is related to the time period \( T \) by the formula:
\[ \omega = \frac{2\pi}{T} \]
For the new time period \( T' \):
\[ \omega' = \frac{2\pi}{T'} \]

6. **Set Up the Equation**:
Substituting \( \omega' \) into the equation gives:
\[ 4 \omega = \frac{2\pi}{T'} \]
Substituting \( \omega = \frac{2\pi}{T} \):
\[ 4 \left(\frac{2\pi}{T}\right) = \frac{2\pi}{T'} \]

7. **Simplify and Solve for New Time Period**:
Cancel \( 2\pi \) from both sides:
\[ 4 \cdot \frac{1}{T} = \frac{1}{T'} \]
Rearranging gives:
\[ T' = \frac{T}{4} \]

8. **Substituting the Original Time Period**:
Since the original time period \( T \) is 24 hours:
\[ T' = \frac{24 \text{ hours}}{4} = 6 \text{ hours} \]

### Final Answer:
The new length of the day would be **6 hours**.

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