A bullet of mass $2\, gm$ is having a charge of $2\,\mu C$. Through what potential difference must it be accelerated, starting from rest, to acquire a speed of $10\,m/s$
AIPMT 2004, Easy
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(b) By using $\frac{1}{2}m{v^2} = QV$
$==>$ $\frac{1}{2} \times 2 \times {10^{ - 3}} \times {(10)^2} = 2 \times {10^{ - 6}}\,V$
$==>$ $V = 50\,kV$
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