Question

A particle with charge $Q$ coulomb, tied at the end of an inextensible string of length $R$ meter, 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 $Mg=\frac{Q^2}{4\pi {\epsilon }_0{R}^2}$. 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

• A. $0$
• B. $2\sqrt{gR}$
• C. $\sqrt{2gR}$
• D. $\sqrt{5gR}$

Answer 1

The correct option is B $2\sqrt{gR}$

$Mg-\frac{K{Q}^2}{R^2}=\frac{m{v}^2}{R}\Rightarrow v=0[\because Mg=\frac{K{Q}^2}{R^2}]$
$\therefore W_g=\Delta KE\Rightarrow mg\left(2R\right)=\frac{1}{2}m{v}_0^2$
$\therefore v_0=2\sqrt{gR}$