Question

A particle $ \, $ $A$ has a charge $q$ and particle $ \, $ $B$ has charge $4q$ with each of them having the mass $m$ . When they are allowed to fall from rest through the same potential difference, the ratio of their speeds $v_{A}:v_{B}$ will be

• A. $4:1$
• B. $1:4$
• C. $1:2$
• D. $2:1$

Answer 1

Answer: (C)

Speed obtained by the particle after falling through a potential difference of $V$ volt is
$v_{A}=\sqrt{\frac{2 V q}{m}} \, \, \ldots (i)$
And $v_{B}= \, \sqrt{\frac{2 V \times 4 q}{m}}$ ...(ii)
Now dividing Eq. (i) by Eq. (ii), we get
$ \, \, \frac{v_{A}}{v_{B}}=\sqrt{\frac{1}{4 \, }}$ = $\frac{1}{2}$
So, $v_{A}:v_{B}=1:2$