If the radius of earth contracts 1/n of its present day value, the length of the day will be approximately

Options

A. `(24)/(n)h`
B. `(24)/(n^(2))h`
C. `24 nh`
D. `24 n^(2)h`

Explanation

To solve the problem, we need to determine how the length of the day changes when the radius of the Earth contracts to \( \frac{1}{n} \) of its present value. We will use the conservation of angular momentum and the relationship between the moment of inertia, angular velocity, and the period of rotation.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- Let the initial radius of the Earth be \( R \).
- The initial time period \( T \) (length of the day) is 24 hours.

2. **Angular Momentum Conservation**:
- The angular momentum \( L \) of a rotating body is given by:
\[ L = I \cdot \omega \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.
- For a solid sphere, the moment of inertia \( I \) is:
\[ I = \frac{2}{5} m R^2 \]
- The angular velocity \( \omega \) is related to the time period \( T \) by:
\[ \omega = \frac{2\pi}{T} \]

3. **New Conditions After Radius Contraction**:
- When the radius contracts to \( R' = \frac{R}{n} \), we need to find the new time period \( T' \).
- The new moment of inertia \( I' \) becomes:
\[ I' = \frac{2}{5} m (R')^2 = \frac{2}{5} m \left(\frac{R}{n}\right)^2 = \frac{2}{5} m \frac{R^2}{n^2} \]

4. **Setting Up the Angular Momentum Equation**:
- By conservation of angular momentum:
\[ L' = L \]
which implies:
\[ I' \cdot \omega' = I \cdot \omega \]
- Substituting the expressions for \( I \) and \( I' \):
\[ \left(\frac{2}{5} m \frac{R^2}{n^2}\right) \cdot \omega' = \left(\frac{2}{5} m R^2\right) \cdot \omega \]
- Simplifying gives:
\[ \frac{R^2}{n^2} \cdot \omega' = R^2 \cdot \omega \]

5. **Solving for the New Angular Velocity**:
- Dividing both sides by \( R^2 \):
\[ \frac{\omega'}{n^2} = \omega \]
- Thus, we find:
\[ \omega' = n^2 \cdot \omega \]

6. **Finding the New Time Period**:
- Since \( \omega' = \frac{2\pi}{T'} \) and \( \omega = \frac{2\pi}{T} \):
\[ \frac{2\pi}{T'} = n^2 \cdot \frac{2\pi}{T} \]
- Cancelling \( 2\pi \) from both sides:
\[ \frac{1}{T'} = n^2 \cdot \frac{1}{T} \]
- Rearranging gives:
\[ T' = \frac{T}{n^2} \]

7. **Substituting the Initial Time Period**:
- Since \( T = 24 \) hours:
\[ T' = \frac{24}{n^2} \text{ hours} \]

8. **Final Answer**:
- Therefore, the length of the day after the contraction of the Earth's radius will be:
\[ T' = \frac{24}{n^2} \text{ hours} \]

### Conclusion:
The correct answer is option 2: \( \frac{24}{n^2} \) hours.