Question

A bullet of mass $2g$ is having a charge of $ 2\mu C $ . Through what potential difference must it be accelerated, starting from rest, to acquire a speed of $10 \,m/s$?

• A. 5 kV
• B. 50 kV
• C. 5 V
• D. 50 V

Answer 1

Answer: (B)

Key Idea : When bullet of mass $m$ and charge $q$ is accelerated through potential difference of $V$ volt, then it attains a kinetic energy equal to $q V$.
Kinetic energy of bullet $=q V$
ie, $ \frac{1}{2} m v^{2} =q V $
$ \Rightarrow V =\frac{m v^{2}}{2 q}$
Given, $m=2 g =2 \times 10^{-3} kg , v=10 \,m / s$
$q=2 \mu C =2 \times 10^{-6} C$
Substituting the values in relation for $V$, we' obtain
$V =\frac{2 \times 10^{-3} \times(10)^{2}}{2 \times 2 \times 10^{-6}}$
$=50 \times 10^{3} V =50 \,kV$