Two rigid bodies `A` and `B` rotate with angular momenta `L_(A)` and `L_(B)` respectively. The moments of inertia of `A` and `B` about the axes of rotation are `I_(A)` and `I_(B)` respectively. If `I_(A)=I_(B)//4` and `L_(A)=5L_(B)`, then the ratio of rotational kinetic energy `K_(A)` of `A` to the rotational kinetic energy `K_(B)` of `B` is given by

Options

A. `(K_(A))/(K_(B))=25/4`
B. `(K_(A))/(K_(B))=5/4`
C. `(K_(A))/(K_(B))=1/4`
D. `(K_(A))/(K_(B))=100`

Explanation

To solve for the ratio of the rotational kinetic energy \( K_A \) of body A to the rotational kinetic energy \( K_B \) of body B, we can follow these steps:

### Step 1: Understand the relationship between angular momentum and rotational kinetic energy
The rotational kinetic energy \( K \) of a rigid body can be expressed in terms of its moment of inertia \( I \) and angular velocity \( \omega \):
\[ K = \frac{1}{2} I \omega^2 \]
We also know that angular momentum \( L \) is related to moment of inertia and angular velocity by:
\[ L = I \omega \]
From this, we can express \( \omega \) in terms of \( L \) and \( I \):
\[ \omega = \frac{L}{I} \]

### Step 2: Substitute \( \omega \) into the kinetic energy formula
Substituting \( \omega \) into the kinetic energy formula gives:
\[ K = \frac{1}{2} I \left( \frac{L}{I} \right)^2 = \frac{L^2}{2I} \]

### Step 3: Write the expressions for \( K_A \) and \( K_B \)
For bodies A and B, we can write:
\[ K_A = \frac{L_A^2}{2I_A} \]
\[ K_B = \frac{L_B^2}{2I_B} \]

### Step 4: Find the ratio \( \frac{K_A}{K_B} \)
The ratio of the kinetic energies is:
\[ \frac{K_A}{K_B} = \frac{\frac{L_A^2}{2I_A}}{\frac{L_B^2}{2I_B}} = \frac{L_A^2}{L_B^2} \cdot \frac{I_B}{I_A} \]

### Step 5: Substitute the given values
From the problem, we know:
- \( I_A = \frac{I_B}{4} \)
- \( L_A = 5 L_B \)

Substituting these values into the ratio:
\[ \frac{K_A}{K_B} = \frac{(5L_B)^2}{(L_B)^2} \cdot \frac{I_B}{\frac{I_B}{4}} = \frac{25L_B^2}{L_B^2} \cdot 4 \]

### Step 6: Simplify the expression
The \( L_B^2 \) terms cancel out:
\[ \frac{K_A}{K_B} = 25 \cdot 4 = 100 \]

### Final Answer
Thus, the ratio of the rotational kinetic energy \( K_A \) of body A to the rotational kinetic energy \( K_B \) of body B is:
\[ \frac{K_A}{K_B} = 100 \]