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Two identical balls having a mass m and small radii and having equal charge q are suspendended from common point by two insulating strings of equal length l . In equilibrium each string makes an angle $\;\theta\;$ with the vertical . If $\;\theta=45^{0}\;find\;l^2$ in terms of q and m

$(a)\;\large\frac{k\;q^2}{2\;mg}\qquad(b)\;\large\frac{k\;q^2}{4\;mg}\qquad(c)\;\large\frac{k\;q^2}{mg}\qquad(d)\;None$

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Answer : (a) $\;\large\frac{k\;q^2}{2\;mg}$
Explanation :
$From\;F.B.D\quad\;T\;cos\;\theta=mg----(1)$
$T\;sin\;\theta=\large\frac{k\;q^2}{4\;l^2\;sin^2\;\theta}----(2)$
From (1) and (2)
$l^2=\large\frac{k\;q^2\;cos\;\theta}{4\;mg\;sin^3\;\theta}$
$l^2=\large\frac{k\;q^2}{2\;mg}\;.$
answered Feb 4, 2014 by yamini.v
 

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