Figure shows a ball having a charge q fixed at a point A. Two identical balls of mass m having charge +q and –q are attached to the end of a light rod of length 2a. The system is released from the situation shown in figure. Find the angular velocity of the rod when the rod turns through 90º.

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$\frac{q}{\sqrt{6 {πε}_{0} {ma}^{3}}}$

$\frac{\sqrt{2} q}{3 {πε}_{0} {ma}^{3}}$

$\frac{q}{\sqrt{3 {πε}_{0} {ma}^{3}}}$

$\frac{\sqrt{2} q}{4 {πε}_{0} {ma}^{3}}$

answer is C.

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Detailed Solution

Initial potential energy = 0

Final potential energy = $\frac{- {kq}^{2}}{a} + \frac{{kq}^{2}}{3 a} = \frac{- 2 {kq}^{2}}{3 a}$

As the total mechanical energy is conserved.

$ΔU + ΔKE = 0$

$\Rightarrow \frac{1}{2} (2 {ma}^{2}) ω^{2} = \frac{2 {kq}^{2}}{3 a}$

$ω = \sqrt{\frac{2 {kq}^{2}}{3 {ma}^{3}}} = \sqrt{\frac{q}{6 {πε}_{0} {ma}^{3}}}$

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