The rotational kinetic energy of a body is E. In the absence of external torque, the mass of the body is halved and its radius of gyration is doubled. Its rotational kinetic energy is

Options

A. 2 E
B. E/2
C. E
D. E/4

Explanation

To solve the problem, we need to analyze the changes in the rotational kinetic energy of a body when its mass is halved and its radius of gyration is doubled.

### Step-by-Step Solution:

1. **Understanding Rotational Kinetic Energy**:
The rotational kinetic energy (KE) of a body is given by the formula:
\[ KE = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

2. **Initial Conditions**:
Let the initial mass of the body be \( m \) and the initial radius of gyration be \( k \). The moment of inertia \( I \) can be expressed as:
\[ I = m k^2 \]
Therefore, the initial rotational kinetic energy \( E \) can be written as:
\[ E = \frac{1}{2} (m k^2) \omega^2 \]

3. **Changes in Mass and Radius of Gyration**:
According to the problem, the mass is halved:
\[ m' = \frac{m}{2} \]
and the radius of gyration is doubled:
\[ k' = 2k \]
The new moment of inertia \( I' \) is then:
\[ I' = m' (k')^2 = \frac{m}{2} (2k)^2 = \frac{m}{2} \cdot 4k^2 = 2m k^2 \]

4. **Conservation of Angular Momentum**:
Since there is no external torque, angular momentum is conserved:
\[ I \omega = I' \omega' \]
Substituting the values we have:
\[ (m k^2) \omega = (2m k^2) \omega' \]
Simplifying gives:
\[ \omega' = \frac{\omega}{2} \]

5. **Calculating New Rotational Kinetic Energy**:
Now we can find the new rotational kinetic energy \( E' \):
\[ E' = \frac{1}{2} I' \omega'^2 \]
Substituting \( I' \) and \( \omega' \):
\[ E' = \frac{1}{2} (2m k^2) \left(\frac{\omega}{2}\right)^2 = \frac{1}{2} (2m k^2) \cdot \frac{\omega^2}{4} \]
Simplifying this expression:
\[ E' = \frac{2m k^2 \omega^2}{8} = \frac{m k^2 \omega^2}{4} \]
Since \( E = \frac{1}{2} m k^2 \omega^2 \), we can relate \( E' \) to \( E \):
\[ E' = \frac{E}{4} \]

### Final Answer:
The new rotational kinetic energy \( E' \) is:
\[ E' = \frac{E}{4} \]