The angular momentum of a body is given by `L=5t^2+2t+1kg m^2/s`.The torque acting on the body at `t=1`s is

Options

A. `12N-m`
B. 10N-m
C. 4N-m
D. Zero

Answer

To solve the problem, we need to find the torque acting on the body at \( t = 1 \) second, given the angular momentum \( L \) as a function of time \( t \).

### Step-by-Step Solution:

1. **Identify the given angular momentum function**:
The angular momentum \( L \) is given by:
\[ L = 5t^2 + 2t + 1 \quad \text{(in kg m}^2/\text{s)} \]

2. **Recall the relationship between torque and angular momentum**:
Torque \( \tau \) is defined as the rate of change of angular momentum with respect to time:
\[ \tau = \frac{dL}{dt} \]

3. **Differentiate the angular momentum function**:
We need to differentiate \( L \) with respect to \( t \):
\[ \frac{dL}{dt} = \frac{d}{dt}(5t^2 + 2t + 1) \]
Using the power rule of differentiation:
- The derivative of \( 5t^2 \) is \( 10t \).
- The derivative of \( 2t \) is \( 2 \).
- The derivative of a constant (1) is \( 0 \).

Thus, we have:
\[ \frac{dL}{dt} = 10t + 2 \]

4. **Substitute \( t = 1 \) into the derivative**:
Now, we will find the torque at \( t = 1 \) second:
\[ \tau = 10(1) + 2 = 10 + 2 = 12 \quad \text{(in N m)} \]

5. **Conclusion**:
Therefore, the torque acting on the body at \( t = 1 \) second is:
\[ \tau = 12 \, \text{N m} \]

### Final Answer:
The torque acting on the body at \( t = 1 \) second is \( 12 \, \text{N m} \).