$(a)\;\sqrt{\large\frac{kQq}{2mR}}\qquad(b)\;\sqrt{\large\frac{kQq}{4mR}}\qquad(c)\;\sqrt{\large\frac{2kQq}{4mR}}\qquad(d)\;\sqrt{\large\frac{kQq}{mR}}$

Answer : (d) $\;\sqrt{\large\frac{kQq}{mR}}$

Explanation :

Work done by electric field on particle = Charge in kinetic energy of particle

Electric field at a distance r from the centre is E = $\;\large\frac{kQr}{R^3}$

Work done by electric field from r to rdr is

$dW=qE dr$

$W=\int_{R}^{0}\;qE\; dr$

$W=\large\frac{kqQ}{R^3}\;\int_{R}^{0}\;r\;dr$

$W=-\large\frac{kqQ}{2R^3} \;R^2=-\large\frac{kqQ}{2R}$

For $\;v_{0}\;$ to be minimum velocity on surface velocity at centre should be zero

Therefore

$\bigtriangleup K.E=\large\frac{1}{2} m (0)^2 - \large\frac{1}{2} mv_{0}^2=-\large\frac{1}{2} mv_{0}^2$

$W=\bigtriangleup K.E$

$-\large\frac{kqQ}{2R}=-\large\frac{1}{2}\;mv_{0}^2$

$v_{0}=\sqrt{\large\frac{kqQ}{mR}}$

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