The momentum of a moving particle given by `p=tl nt`. Net force acting on this particle is defined by equation `F=(dp)/(dt)`. The net force acting on the particle is zero at time

Options

A. `t=0`
B. `t=1/e`
C. `t=1/e^(2)`
D. None of these

Explanation

To solve the problem, we need to find the time at which the net force acting on the particle is zero. The momentum of the particle is given by the equation:

\[ p = t \log t \]

The net force \( F \) acting on the particle is defined by the equation:

\[ F = \frac{dp}{dt} \]

### Step 1: Differentiate the momentum with respect to time
We start by differentiating the momentum \( p \) with respect to time \( t \):

\[ \frac{dp}{dt} = \frac{d}{dt}(t \log t) \]

Using the product rule of differentiation, where \( u = t \) and \( v = \log t \):

\[ \frac{dp}{dt} = u \frac{dv}{dt} + v \frac{du}{dt} \]

Calculating \( \frac{du}{dt} \) and \( \frac{dv}{dt} \):

- \( \frac{du}{dt} = 1 \)
- \( \frac{dv}{dt} = \frac{1}{t} \)

Now applying the product rule:

\[ \frac{dp}{dt} = t \cdot \frac{1}{t} + \log t \cdot 1 \]
\[ \frac{dp}{dt} = 1 + \log t \]

### Step 2: Set the force equal to zero
The net force \( F \) is given by:

\[ F = \frac{dp}{dt} \]

To find when the force is zero, we set the expression for \( \frac{dp}{dt} \) equal to zero:

\[ 1 + \log t = 0 \]

### Step 3: Solve for \( t \)
Now we solve the equation:

\[ \log t = -1 \]

To eliminate the logarithm, we exponentiate both sides:

\[ t = e^{-1} \]

This simplifies to:

\[ t = \frac{1}{e} \]

### Conclusion
Thus, the net force acting on the particle is zero at time \( t = \frac{1}{e} \).