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A particle with charge $Q$ coulomb, tied at the end of an inextensible string of length $R$ metre, revolves in a vertical plane. At the centre of the circular trajectory, there is a fixed charge of magnitude $Q$ coulomb. The mass of the moving charge $M$ is such that $M g=\frac{Q^{2}}{4 \pi \in_{0} R^{2}}$. If at the highest position of the particle, the tension of the string just vanishes, the horizontal velocity at the lowest point has to be
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Verified Answer
The correct answer is:
$2 \sqrt{g R}$
According to the question, when tension in the string is zero.
As $T=0$ $M g-\frac{K Q^{2}}{R^{2}}=\frac{m v^{2}}{R} \quad\left(k=\frac{1}{4 \pi \varepsilon_{0}}\right)$
$\frac{m v^{2}}{R}=$ centripetal force required to keep the body in a circular path
$\Rightarrow \quad v=0$$\quad\left[\because M g=\frac{k Q^{2}}{R^{2}}\right]$
So the required work done to keep charge in vertical motion
$\therefore \quad W_{g}=\Delta K E=$ change in $\mathrm{K.E.}$
$\Rightarrow \quad m g(2 R)=\frac{1}{2} m v_{0}^{2}$
\begin{array}{l}
v_{0}^{2}=2 g \cdot 2 R=4 g R \\
v_{0}=\sqrt{4 g R}=2 \sqrt{g R}
\end{array}
As $T=0$ $M g-\frac{K Q^{2}}{R^{2}}=\frac{m v^{2}}{R} \quad\left(k=\frac{1}{4 \pi \varepsilon_{0}}\right)$
$\frac{m v^{2}}{R}=$ centripetal force required to keep the body in a circular path
$\Rightarrow \quad v=0$$\quad\left[\because M g=\frac{k Q^{2}}{R^{2}}\right]$
So the required work done to keep charge in vertical motion
$\therefore \quad W_{g}=\Delta K E=$ change in $\mathrm{K.E.}$
$\Rightarrow \quad m g(2 R)=\frac{1}{2} m v_{0}^{2}$
\begin{array}{l}
v_{0}^{2}=2 g \cdot 2 R=4 g R \\
v_{0}=\sqrt{4 g R}=2 \sqrt{g R}
\end{array}
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