If radius if earth is reduced to half without changing its mass, then match the following columns
`{:(,"Column-I",,"Column-II"),("(A)","Angular momentum of earth","(p)","Will become two times"),("(B)","Time period of rotation of earth","(q)","Will become four times"),("(C)","Rotational kinetic energy of earth","(r)","Will remain constant"),("(D)",,"(s)","None"):}`

Explanation

To solve the problem, we need to analyze how the angular momentum, time period of rotation, and rotational kinetic energy of the Earth change when its radius is reduced to half while keeping its mass constant.

### Step-by-Step Solution:

1. **Angular Momentum of Earth (L)**:
- The angular momentum \( L \) is given by the formula:
\[ L = I \cdot \omega \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.
- The moment of inertia \( I \) for a solid sphere is:
\[ I = \frac{2}{5} m r^2 \]
- If the radius is reduced to half, the new radius \( r_2 = \frac{r}{2} \):
\[ I_2 = \frac{2}{5} m \left(\frac{r}{2}\right)^2 = \frac{2}{5} m \cdot \frac{r^2}{4} = \frac{1}{10} m r^2 \]
- By conservation of angular momentum:
\[ L_1 = L_2 \implies I_1 \cdot \omega_1 = I_2 \cdot \omega_2 \]
- Substituting the values:
\[ \frac{2}{5} m r^2 \cdot \omega = \frac{1}{10} m r^2 \cdot \omega_2 \]
- Canceling \( m r^2 \) from both sides:
\[ \frac{2}{5} \cdot \omega = \frac{1}{10} \cdot \omega_2 \implies \omega_2 = 4 \cdot \omega \]
- Since angular momentum is conserved, it remains constant:
\[ L \text{ will remain constant.} \]
- Thus, **(A) matches with (r)**.

2. **Time Period of Rotation (T)**:
- The time period \( T \) is related to angular velocity by:
\[ T = \frac{2\pi}{\omega} \]
- The new angular velocity \( \omega_2 = 4 \cdot \omega \):
\[ T_2 = \frac{2\pi}{\omega_2} = \frac{2\pi}{4\omega} = \frac{1}{4} T_1 \]
- This indicates that the time period of rotation will decrease:
\[ T \text{ will become } \frac{1}{4} \text{ of the original time period.} \]
- Thus, **(B) matches with (s)** (none of the options).

3. **Rotational Kinetic Energy (K)**:
- The rotational kinetic energy is given by:
\[ K = \frac{1}{2} I \omega^2 \]
- Initial kinetic energy:
\[ K_1 = \frac{1}{2} \cdot \frac{2}{5} m r^2 \cdot \omega^2 = \frac{1}{5} m r^2 \cdot \omega^2 \]
- Final kinetic energy:
\[ K_2 = \frac{1}{2} \cdot \frac{1}{10} m r^2 \cdot (4\omega)^2 = \frac{1}{2} \cdot \frac{1}{10} m r^2 \cdot 16\omega^2 = \frac{8}{10} m r^2 \cdot \omega^2 = \frac{4}{5} m r^2 \cdot \omega^2 \]
- Thus, the final kinetic energy is:
\[ K_2 = 4 \cdot K_1 \]
- Therefore, **(C) matches with (q)**.

### Final Matching:
- (A) Angular momentum of earth → (r) Will remain constant
- (B) Time period of rotation of earth → (s) None
- (C) Rotational kinetic energy of earth → (q) Will become four times