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 vA:vB will be

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 vA:vB will be (A) 4:1 (B) 1:4 (C) 1:2 (D) 2:1

Tardigrade Admin · Accepted Answer

Speed obtained by the particle after falling through a potential difference of

V

volt is

v_{A} = \frac{2 V q}{m} \dots (i)

And

v_{B} = \frac{2 V \times 4 q}{m}

...(ii)
Now dividing Eq. (i) by Eq. (ii), we get

\frac{v _{A}}{v _{B}} = \frac{1}{4}

=

\frac{1}{2}

So,

v_{A} : v_{B} = 1 : 2

Q. A particle $A$ has a charge $q$ and particle $B$ has charge $4 q$ 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

$4 : 1$

$1 : 4$

$1 : 2$

$2 : 1$

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

Q. 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 vA​:vB​ will be

4:1

1:4

1:2

2:1