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(A) Let the linear charge density of the wire be $\lambda$. The potential difference between two points at distances $r_1$ and $r_2$ from a long charg

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$\sqrt{3}u$

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(A) Let the linear charge density of the wire be $\lambda$. The potential difference between two points at distances $r_1$ and $r_2$ from a long charged wire is given by $V_1 - V_2 = \frac{\lambda}{2\pi\varepsilon_0} \ln\left(\frac{r_2}{r_1}\right)$.Applying the principle of conservation of energy between point $A$ (at distance $l$) and point $B$ (at distance $2l$):$qV_A + \frac{1}{2}mu^2 = qV_B + \frac{1}{2}m(\sqrt{2}u)^2$$q(V_A - V_B) = \frac{1}{2}m(2u^2 - u^2) = \frac{1}{2}mu^2$$q \frac{\lambda}{2\pi\varepsilon_0} \ln\left(\frac{2l}{l}\right) = \frac{1}{2}mu^2$$q \frac{\lambda}{2\pi\varepsilon_0} \ln(2) = \frac{1}{2}mu^2 \quad \dots(i)$Now, applying energy conservation between point $A$ (at distance $l$) and point $C$ (at distance $4l$):$qV_A + \frac{1}{2}mu^2 = qV_C + \frac{1}{2}mv^2$$\frac{1}{2}mv^2 = q(V_A - V_C) + \frac{1}{2}mu^2$$\frac{1}{2}mv^2 = q \frac{\lambda}{2\pi\varepsilon_0} \ln\left(\frac{4l}{l}\right) + \frac{1}{2}mu^2$$\frac{1}{2}mv^2 = q \frac{\lambda}{2\pi\varepsilon_0} \ln(4) + \frac{1}{2}mu^2$Since $\ln(4) = 2\ln(2)$, we have:$\frac{1}{2}mv^2 = 2 \left(q \frac{\lambda}{2\pi\varepsilon_0} \ln(2)\right) + \frac{1}{2}mu^2$Substituting from equation $(i)$:$\frac{1}{2}mv^2 = 2 \left(\frac{1}{2}mu^2\right) + \frac{1}{2}mu^2 = \frac{3}{2}mu^2$$v^2 = 3u^2 \Rightarrow v = \sqrt{3}u$.

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