If the mass of earth and radius suddenly become `2` times and `1//4th` of the present value, the length of the day becomes

Options

A. `24 h`
B. `6h`
C. `3//2 h`
D. `3h`

Explanation

To solve the problem, we need to analyze the changes in the Earth's mass and radius and how they affect the length of a day.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The mass of the Earth (M) becomes 2 times its current mass: \( M_2 = 2M_1 \).
- The radius of the Earth (R) becomes \( \frac{1}{4} \) of its current radius: \( R_2 = \frac{1}{4}R_1 \).
- We need to find the new length of the day (T).

2. **Conservation of Angular Momentum**:
- The angular momentum (L) of a rotating body is conserved if no external torques act on it.
- The angular momentum can be expressed as:
\[ L = I \cdot \omega \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

3. **Moment of Inertia for a Solid Sphere**:
- The moment of inertia \( I \) of a solid sphere is given by:
\[ I = \frac{2}{5} M R^2 \]
- For the initial state (before the changes), we have:
\[ I_1 = \frac{2}{5} M_1 R_1^2 \]
- For the new state (after the changes), we have:
\[ I_2 = \frac{2}{5} M_2 R_2^2 = \frac{2}{5} (2M_1) \left(\frac{1}{4}R_1\right)^2 = \frac{2}{5} (2M_1) \left(\frac{1}{16}R_1^2\right) = \frac{2M_1 R_1^2}{80} = \frac{M_1 R_1^2}{40} \]

4. **Setting Up the Angular Momentum Conservation Equation**:
- Since angular momentum is conserved:
\[ I_1 \omega_1 = I_2 \omega_2 \]
- Substituting the expressions for \( I_1 \) and \( I_2 \):
\[ \frac{2}{5} M_1 R_1^2 \omega_1 = \frac{M_1 R_1^2}{40} \omega_2 \]

5. **Cancelling Common Factors**:
- We can cancel \( M_1 \) and \( R_1^2 \) from both sides:
\[ \frac{2}{5} \omega_1 = \frac{1}{40} \omega_2 \]

6. **Solving for Angular Velocities**:
- Rearranging gives:
\[ \omega_2 = \frac{2}{5} \cdot 40 \cdot \omega_1 = 16 \omega_1 \]

7. **Finding the New Period of Rotation**:
- The period \( T \) is related to angular velocity \( \omega \) by:
\[ T = \frac{2\pi}{\omega} \]
- The initial period \( T_1 \) (length of the day) is 24 hours:
\[ T_1 = \frac{2\pi}{\omega_1} \]
- The new period \( T_2 \) is:
\[ T_2 = \frac{2\pi}{\omega_2} = \frac{2\pi}{16 \omega_1} = \frac{T_1}{16} = \frac{24 \text{ hours}}{16} = 1.5 \text{ hours} = \frac{3}{2} \text{ hours} \]

8. **Final Answer**:
- The new length of the day is \( \frac{3}{2} \) hours.