A gun applies a force F on a bullet which is given by `F=(100-0.5xx10^(5)t)N`. The bullet emerges out with speed 400 m/s. then find out the impulsive exerted till force on bullet becomes zero.

Options

A. 0.2N-s
B. 0.3N-s
C. 0.1N-s
D. 0.4N-s

Explanation

To solve the problem, we need to find the impulse exerted on the bullet until the force becomes zero. The force acting on the bullet is given by:

\[ F(t) = 100 - 0.5 \times 10^5 t \]

### Step 1: Determine when the force becomes zero
We need to find the time \( t \) when the force \( F(t) \) is equal to zero.

Setting the force equation to zero:

\[ 100 - 0.5 \times 10^5 t = 0 \]

Solving for \( t \):

\[ 0.5 \times 10^5 t = 100 \]

\[ t = \frac{100}{0.5 \times 10^5} = \frac{100}{50000} = 2 \times 10^{-3} \text{ seconds} \]

### Step 2: Calculate the impulse
Impulse \( I \) is defined as the integral of force over time:

\[ I = \int_0^{t_f} F(t) \, dt \]

Where \( t_f = 2 \times 10^{-3} \) seconds. Thus, we need to evaluate:

\[ I = \int_0^{2 \times 10^{-3}} (100 - 0.5 \times 10^5 t) \, dt \]

### Step 3: Evaluate the integral
We can split the integral into two parts:

\[ I = \int_0^{2 \times 10^{-3}} 100 \, dt - \int_0^{2 \times 10^{-3}} 0.5 \times 10^5 t \, dt \]

Calculating the first integral:

\[ \int_0^{2 \times 10^{-3}} 100 \, dt = 100 \cdot t \bigg|_0^{2 \times 10^{-3}} = 100 \cdot (2 \times 10^{-3}) = 0.2 \text{ Ns} \]

Calculating the second integral:

\[ \int_0^{2 \times 10^{-3}} 0.5 \times 10^5 t \, dt = 0.5 \times 10^5 \cdot \frac{t^2}{2} \bigg|_0^{2 \times 10^{-3}} = 0.5 \times 10^5 \cdot \frac{(2 \times 10^{-3})^2}{2} \]

Calculating \( (2 \times 10^{-3})^2 = 4 \times 10^{-6} \):

\[ = 0.5 \times 10^5 \cdot \frac{4 \times 10^{-6}}{2} = 0.5 \times 10^5 \cdot 2 \times 10^{-6} = 1 \times 10^{-1} \text{ Ns} \]

### Step 4: Combine the results
Now, substituting back into the impulse equation:

\[ I = 0.2 - 0.1 = 0.1 \text{ Ns} \]

### Final Answer
Thus, the impulse exerted until the force on the bullet becomes zero is:

\[ \boxed{0.1 \text{ Ns}} \]