$(a)\;\sqrt{26} m/s\qquad(b)\;\sqrt{31} m/s\qquad(c)\;6 m/s\qquad(d)\;zero$

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Answer : (b) $\;\sqrt{31} m/s$

Explanation :

Let the speed of the particle at C = v

Then ,

Centripetal force =$\;mg cos 30^{0}-qE cos 60^{0} $

$=\large\frac{10 \sqrt{3}}{2}-\large\frac{10}{2}=5\;\sqrt{3}-5$

and $\;F_{C}=\large\frac{mV^2}{l}=5 sqrt{5}-5$

$V^2=5 \sqrt{3} -5---(1)$

By energy conseration :

$\bigtriangleup K.E=\large\frac{1}{2}\;mV^2-\large\frac{1}{2}\;mu^2$

Work done by gravity = $\;-mgl \;(1+sin \theta)_{| \theta=60^{0}}$

Work done by electric field =$\;qEl cos \theta$

Work done by all force = $\;\bigtriangleup K.E$

$-mgl\;(1+sin 60^{0})+qEl cos 60^{0}=\large\frac{1}{2} mV^2-\large\frac{1}{2} mu^2$

$\large\frac{V^2-u^2}{2}=-10\;(1+\large\frac{sqrt{3}}{2})+\large\frac{10\times1}{2}$

$V^2-u^2=2\;(-5-5 \sqrt{3})$

$u^2=V^2+(5+5 \sqrt{3})\;2$

$u^2=5 sqrt{3}-5+10+10 \sqrt{3}$

$u^2=5\;(1+3 \sqrt{3})$

$u^2=5\;(1+3\times1.73)$

$u^2 \approx 31$

$u=\sqrt{31}$

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