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Two particles A and B having same mass have charge $+q$ and $+4 \mathrm{q}$ respectively. When they are allowed to fall from rest through same electric potential difference, of ratio of their speeds ' $V_A$ ' to ' $V_B$ ' will become
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The correct answer is:
1:2
If $\mathrm{V}$ is the potential difference then
$$
\begin{aligned}
& \frac{1}{2} \mathrm{~m} v_{\mathrm{A}}^2=\mathrm{qV} \\
& \text { and } \frac{1}{2} \mathrm{~m} v_{\mathrm{B}}^2=4 \mathrm{qV} \\
& \therefore \frac{v_{\mathrm{A}}^2}{v_{\mathrm{B}}^2}=\frac{1}{4} \\
& \therefore \frac{v_{\mathrm{A}}}{v_{\mathrm{B}}}=\frac{1}{2}
\end{aligned}
$$
$$
\begin{aligned}
& \frac{1}{2} \mathrm{~m} v_{\mathrm{A}}^2=\mathrm{qV} \\
& \text { and } \frac{1}{2} \mathrm{~m} v_{\mathrm{B}}^2=4 \mathrm{qV} \\
& \therefore \frac{v_{\mathrm{A}}^2}{v_{\mathrm{B}}^2}=\frac{1}{4} \\
& \therefore \frac{v_{\mathrm{A}}}{v_{\mathrm{B}}}=\frac{1}{2}
\end{aligned}
$$
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