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# A simple pendulum with a bob of mass m=1 kg , charge $5\;mu C \;$ and string length l=1m is given a horizontal velocity u in a uniform electric field $\;E=2\times10^{6} V/m\;$ at its bottom most point as shown in figure . It is given that the speed u is such that the particle leaves the circle at point C . Find approx the speed u in m/s (Take g=10 m/s^2)

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

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A)
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 = $\;\bigtriangleup K.E$
$-mgl\;(1+sin 60^{0})+qEl cos 60^{0}=\large\frac{1}{2} mV^2-\large\frac{1}{2} mu^2$
$\large\frac{V^2-u^2}{2}=-10\;(1+\large\frac{sqrt{3}}{2})+\large\frac{10\times1}{2}$
$V^2-u^2=2\;(-5-5 \sqrt{3})$
$u^2=V^2+(5+5 \sqrt{3})\;2$
$u^2=5 sqrt{3}-5+10+10 \sqrt{3}$
$u^2=5\;(1+3 \sqrt{3})$
$u^2=5\;(1+3\times1.73)$
$u^2 \approx 31$
$u=\sqrt{31}$