A simple pendulum with a bob of mass $m = 1\ kg$ , charge $q = 5\mu C$ and string length $l = 1\ m$ is given a horizontal velocity $u$ in a uniform electric field $E = 2 × 10^6\ V/m$ at its bottom most point $A$ , as shown in figure. It is given a speed $u$ such that the particle leave the circular path at its topmost point $C$ . Find the speed $u$ . (Take $g = 10\ m/s^2$ )
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By applying condition for circular motion net force toward centre at point $\mathrm{C}$ should be equal to $\mathrm{mv}^{2} / \mathrm{R}$
By applying work energy theoram at point $\mathrm{A}$ and $\mathrm{C}$
$\frac{1}{2} \mathrm{mv}^{2}-\frac{1}{2} \mathrm{mu}^{2}=$ work done by all the forces
$\Rightarrow \frac{1}{2} \mathrm{mv}^{2}-\frac{1}{2} \mathrm{mRg}=\mathrm{W}_{\text {electric field }}+\mathrm{W}_{\text {gravity }}$
$\Rightarrow \frac{1}{2} \mathrm{mv}^{2}-\frac{1}{2} \mathrm{mRg}=0-2 \mathrm{mgh}$
$\Rightarrow \mathrm{v}=\sqrt{5 \mathrm{Rg}}=\sqrt{5 \times 1 \times 10}=\sqrt{50} \mathrm{\,m} / \mathrm{s}$
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