$(a)\;2 \sqrt{2} m/s\qquad(b)\;3 \sqrt{3} m/s\qquad(c)\;2 m/s\qquad(d)\;3 m/s$

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Answer : (d) 3 m/s

Explanation :

The least value of kinetic energy of the particle at infinity should be enough to take the particle upto $x=+\sqrt{\large\frac{5}{2}}\;$ because

at $x=+\sqrt{\large\frac{5}{2}}\;,E=0\;$ Electrostatic force on $\;q_{0}\;$ is zero

for $\;x > \sqrt{\large\frac{5}{2}}\;,E\;$ is repulsive (towards +ve x-axis)

for $\;x < \sqrt{\large\frac{5}{2}}\;,E\;$ is attractive (towards _ve x-axis)

Now potential at $\;\sqrt{\large\frac{5}{2}} m\;$ is

$V=1.8\times10^{4}\;[\large\frac{8}{(\large\frac{27}{2}+\large\frac{5}{2})^{\large\frac{1}{2}}}-\large\frac{1}{(\large\frac{3}{2}+\large\frac{5}{2})^{\large\frac{1}{2}}}]$

$V=1.8\times1o^{4}\;[\large\frac{8}{4}-\large\frac{1}{2}]$

$V=1.8\times10^{4}\times\large\frac{3}{2}$

$V=2.7\times10^{4}V$

Applying energy conservation at $\;x=\infty\;and\;$ $\;x=\sqrt{\large\frac{5}{2}} m$

$\large\frac{1}{2}\;mv_{0}^{2}=q_{0}V$

$v_{0}=\sqrt{\large\frac{2q_{0}V}{m}}$

$v_{0}=\sqrt{\large\frac{2\times10^{-7}\times2.7\times10^{4}}{6\times10^{-4}}}$

$v_{0}=3 m/s$

Minimum value of $\;v_{0}\;$ is 3 m/s

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