COMEDK · Physics · 17. Electrostatics
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
- A \(2: 1\)
- B \(1: 2\)
- C \(1: 4\)
- D \(4: 1\)
Answer & Solution
Correct Answer
(B) \(1: 2\)
Step-by-step Solution
Detailed explanation
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 \(V\) ] \(\Rightarrow \quad v \propto \sqrt{q}\)…
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