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A particle $A$ has charge $+q$ and particle $B$ has charge $+4 q$, each of them having the same mass $m$. When allowed to fall from rest through the same electrical potential difference, then the ratio of their speeds will become
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Verified Answer
The correct answer is:
$1: 2$
Given, $m_{1}=m_{2}=m$
$$
\begin{aligned}
&q_{1}=q \\
&q_{2}=4 q
\end{aligned}
$$
We know that
$E=\frac{1}{2} m v^{2}=q V$
$\Rightarrow \quad v^{2} \propto q$
[For same value of potential difference $V$ ]
$\Rightarrow \quad v \propto \sqrt{q}$
$\Rightarrow \quad \frac{v_{1}}{v_{2}}=\sqrt{\frac{q_{1}}{q_{2}}}$
$=\sqrt{\frac{q}{4 q}}=\frac{1}{2}$
$\Rightarrow \quad v_{1}: v_{2}=1: 2$
$$
\begin{aligned}
&q_{1}=q \\
&q_{2}=4 q
\end{aligned}
$$
We know that
$E=\frac{1}{2} m v^{2}=q V$
$\Rightarrow \quad v^{2} \propto q$
[For same value of potential difference $V$ ]
$\Rightarrow \quad v \propto \sqrt{q}$
$\Rightarrow \quad \frac{v_{1}}{v_{2}}=\sqrt{\frac{q_{1}}{q_{2}}}$
$=\sqrt{\frac{q}{4 q}}=\frac{1}{2}$
$\Rightarrow \quad v_{1}: v_{2}=1: 2$
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