A small ball of mass $2 x {10}^{-3} kg$ having a charge of 1 $\mu$C is suspended by a string of length 0.8m. Another identical ball having the same charge is kept at the point of suspension. Determine the minimum horizontal velocity in $cm{s}^{-1}$ which should be imparted to the lower ball so that it can make complete revolution. (take g= 10ms^-2 and $\sqrt{\frac{11}{2}}=2.3$)

For the ball just to complete the circle, the tension must vanish at the topmost point i.e $T_2=0$

From Newton Second Law,

$T_2+mg-\frac{q^2}{4\pi {\epsilon }_0{\mathcal{l}}^2}=\frac{m{v}^2}{\mathcal{l}}$

At the topmost point $T_2=0$

$mg-\frac{q^2}{4\pi {\epsilon }_0{\mathcal{l}}^2}=\frac{m{v}^2}{\mathcal{l}}$....(1)

By Law of Conservation of Energy

$\left(\begin{array}{l}Energyatthe\\ lowestpo\mathrm{int}L\end{array}\right)=\left(\begin{array}{l}Energyatthe\\ highestpo\mathrm{int}H\end{array}\right)$

$\Rightarrow \frac{1}{2}m{u}^2=\frac{1}{2}m{v}^2+mg\left(2\mathcal{l}\right)$

$\Rightarrow u^2-v^2=4g\mathcal{l}$.....(2)

$Fromequation\left(1\right)v^2=g\mathcal{l}-\frac{q^2}{4\pi {\epsilon }_{0}m\mathcal{l}}$
$u^2=5g\mathcal{l}-\frac{q^2}{4\pi {\epsilon }_{0}m\mathcal{l}}$

$\Rightarrow u=\sqrt{\frac{275}{8}}=\frac{5}{2}\sqrt{\frac{11}{2}}=5.75m{s}^{-1}$

$\Rightarrow u=575cm{s}^{-1}$