The moment of an inertia about an axis of a body which is rotating with angular velocity `"1rad s"^(-1)` is numerically equal to

Options

A. one - fourth of its rotational kinetic energy
B. half of the rotational kinetic energy
C. rotational kinetic energy
D. twice the rotational kinetic energy

Explanation

To solve the problem, we need to find the moment of inertia (I) of a body that is rotating with an angular velocity (ω) of 1 rad/s. We will use the formula for rotational kinetic energy (K.E.) to derive the moment of inertia.

### Step-by-Step Solution:

1. **Identify the formula for rotational kinetic energy**:
The rotational kinetic energy (K.E.) of a rotating body is given by the formula:
\[ K.E. = \frac{1}{2} I \omega^2 \]

2. **Substitute the given values**:
We know that the angular velocity (ω) is 1 rad/s. Therefore, substituting this value into the kinetic energy formula:
\[ K.E. = \frac{1}{2} I (1)^2 = \frac{1}{2} I \]

3. **Express moment of inertia in terms of kinetic energy**:
From the equation above, we can express the moment of inertia (I) in terms of the kinetic energy (K.E.):
\[ K.E. = \frac{1}{2} I \implies I = 2 \times K.E. \]

4. **Conclusion**:
The moment of inertia (I) is numerically equal to twice the rotational kinetic energy. Thus, we can conclude:
\[ I = 2 \times K.E. \]

5. **Final answer**:
Since the problem states that the moment of inertia is numerically equal to the kinetic energy, we can say:
\[ I = 2 \times K.E. \quad \text{(numerically)} \]