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At a distance $\ell$ from a uniformly charged long wire, a charged particle is thrown radially outward with a velocity $\mathrm{u}$ in the direction perpendicular to the wire. When the particle reaches a distance $2 \ell$ from the wire its speed is found to be $\sqrt{2} \mathrm{u}$. The magnitude of the velocity, when it is a distance $4 \ell$ away from the wire, is (ignore gravity)
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The correct answer is:
$\sqrt{3} \mathrm{u}$
energy conservation at A \& B
$\begin{array}{l}
\mathrm{qV}_{\Lambda}+\frac{1}{2} \mathrm{mu}^{2}=\mathrm{qV}_{\mathrm{B}}+\frac{1}{2} \mathrm{~m} \times 2 \mathrm{u}^{2} \\
\mathrm{q}\left[\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}\right]=\frac{1}{2} \mathrm{mu}^{2} \\
\mathrm{q} \times \frac{\lambda}{2 \pi \in_{0}} \ln 2=\frac{1}{2} \mathrm{mu}^{2}
\end{array}$
energy conservation at $\mathrm{A} \& \mathrm{C}$
$\begin{array}{l}
\mathrm{qV}_{\mathrm{A}}+\frac{1}{2} \mathrm{mu}^{2}=\mathrm{q} \mathrm{V}_{\mathrm{C}}+\frac{1}{2} \mathrm{mv}^{2} \\
\mathrm{q}\left[\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{C}}\right]+\frac{1}{2} \mathrm{mu}^{2}=\frac{1}{2} \mathrm{mv}^{2} \\
\frac{\mathrm{q} \lambda}{2 \pi \epsilon_{0}} \ln 4+\frac{1}{2} \mathrm{mu}^{2}=\frac{1}{2} \mathrm{mv}^{2} \\
\frac{2 \mathrm{q} \lambda}{2 \pi \in_{0}} \ln 2+\frac{1}{2} \mathrm{mu}^{2}=\frac{1}{2} \mathrm{mv}^{2} \\
\mathrm{mu}^{2}+\frac{1}{2} \mathrm{mu}^{2}=\frac{1}{2} \mathrm{mv}^{2} \\
\frac{3}{2} \mathrm{u}^{2}=\frac{1}{2} \mathrm{v}^{2} \Rightarrow \mathrm{v}=\sqrt{3} \mathrm{u}
\end{array}$
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