Angular impulse J is applied on a system which can rotate about an axis for which its moment of inertia is I. System is initially at rest.

Options

A. Angular momentum of the system becomes J
B. Angular velocity acquired by the system is J/I
C. Kinetic energy acquired by the system is `(J^(2))/(2I)`
D. Angular momentum of the system becomes J/I

Explanation

To solve the problem step by step, we need to analyze the given information and apply the relevant physics concepts.

### Step 1: Understand Angular Impulse
Angular impulse \( J \) is defined as the product of torque \( \tau \) and the time duration \( dt \) over which the torque is applied. Mathematically, it can be expressed as:
\[ J = \tau \cdot dt \]

### Step 2: Relate Torque to Angular Acceleration
Torque \( \tau \) can also be expressed in terms of moment of inertia \( I \) and angular acceleration \( \alpha \):
\[ \tau = I \cdot \alpha \]

### Step 3: Relate Angular Acceleration to Angular Velocity
Since the system starts from rest, the initial angular velocity \( \omega_0 = 0 \). After applying the angular impulse, the final angular velocity is \( \omega \). The change in angular velocity \( d\omega \) is:
\[ d\omega = \omega - \omega_0 = \omega \]

### Step 4: Substitute into Angular Impulse Equation
Substituting the expression for torque into the angular impulse equation gives:
\[ J = I \cdot d\omega \]
Since \( d\omega = \omega \), we can rewrite it as:
\[ J = I \cdot \omega \]

### Step 5: Solve for Angular Velocity
From the equation \( J = I \cdot \omega \), we can solve for \( \omega \):
\[ \omega = \frac{J}{I} \]

### Step 6: Calculate Angular Momentum
The angular momentum \( L \) of the system is given by:
\[ L = I \cdot \omega \]
Substituting \( \omega \) from the previous step:
\[ L = I \cdot \left(\frac{J}{I}\right) = J \]

### Step 7: Calculate Kinetic Energy
The kinetic energy \( K \) of a rotating system is given by:
\[ K = \frac{1}{2} I \omega^2 \]
Substituting \( \omega \) from Step 5:
\[ K = \frac{1}{2} I \left(\frac{J}{I}\right)^2 = \frac{1}{2} I \cdot \frac{J^2}{I^2} = \frac{J^2}{2I} \]

### Summary of Results
1. The angular momentum of the system becomes \( J \).
2. The angular velocity acquired by the system is \( \frac{J}{I} \).
3. The kinetic energy acquired by the system is \( \frac{J^2}{2I} \).

### Conclusion
Based on the analysis, the correct answers to the options provided in the question are:
- Angular momentum becomes \( J \)
- Angular velocity acquired is \( \frac{J}{I} \)
- Kinetic energy acquired is \( \frac{J^2}{2I} \)