The torque applied to a ring revolving about its own axis so as to change its angular momentum by 2 J-s. in 5 s, is

Options

A. `10N-m`
B. `2.5 N-m`
C. `0.1 N-m`
D. `0.4 N-m`

Answer

To find the torque applied to a ring revolving about its own axis that changes its angular momentum by 2 J-s in 5 seconds, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between torque and angular momentum**:
The torque (\( \tau \)) applied to an object is related to the change in angular momentum (\( \Delta L \)) over time (\( \Delta t \)) by the equation:
\[ \tau = \frac{\Delta L}{\Delta t} \]

2. **Identify the given values**:
From the problem, we know:
- Change in angular momentum (\( \Delta L \)) = 2 J-s (joule-seconds)
- Time interval (\( \Delta t \)) = 5 s

3. **Substitute the values into the torque equation**:
Now we can substitute the known values into the torque equation:
\[ \tau = \frac{2 \, \text{J-s}}{5 \, \text{s}} \]

4. **Calculate the torque**:
Performing the calculation gives:
\[ \tau = \frac{2}{5} = 0.4 \, \text{N-m} \]

5. **Conclusion**:
The torque applied to the ring is \( 0.4 \, \text{N-m} \).

### Final Answer:
The torque applied to the ring is \( 0.4 \, \text{N-m} \).

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