MHT CET · Physics · Electrostatics
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
- A 1:2
- B 2:1
- C 1:4
- D 4:1
Answer & Solution
Correct Answer
(A) 1:2
Step-by-step Solution
Detailed explanation
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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