Q107 / 180 · AIIMS 2019 (25 May) · Physics
A gun applies a force $F$ on a bullet which is given by $F=\left(100-0.5 \times 10^5 t\right) \mathrm{N}$. The bullet emerges out with speed $400 \mathrm{~m} \mathrm{~s}^{-1}$. Then find out the impulse exerted till force on bullet becomes zero
- A$0.2 \mathrm{~N} \mathrm{~s}$
- B$0.3 \mathrm{~N} \mathrm{~s}$
- Correct answerC$0.1 \mathrm{~N} \mathrm{~s}$
- D$0.4 \mathrm{Ns}$
Solution
$\begin{aligned} & \text { (c) : Given, force } F=\left(100-0.5 \times 10^5 t\right) \mathrm{N} \\ & 100-0.5 \times 10^5 t=0,100=0.5 \times 10^5 t(\because F=0) \\ & t=2 \times 10^{-3} \mathrm{sec} \\ & \text { Impulse, } I=\int F d t=\int_0^{2 \times 10^{-3}}\left(100-0.5 \times 10^5 t\right) d t \\ & I=\left[100 t-\frac{10^5}{2} \frac{t^2}{2}\right]_0^{2 \times 10^{-3}} \\ & I=\left[100 \times 2 \times 10^{-3}-\frac{10^5}{4} \times 4 \times 10^{-6}\right] \\ & I=10^{-1}=0.1 \mathrm{~N} \mathrm{~s}\end{aligned}$
Hence, the correct answer is option: 'C'.