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

Answer: (B)

According to the question, when tension in the string is zero.

As $T=0$
$Mg-\frac{K{Q}^2}{R^2}=\frac{m{v}^2}{R}(k=\frac{1}{4\pi {\epsilon }_0})$
$\frac{m{v}^2}{R}=$ centripetal force required to keep the body in a circular path
$\Rightarrow v=0\left[\because Mg=\frac{k{Q}^2}{R^2}\right]$
So the required work done to keep charge in vertical motion
$\therefore W_8=\Delta KE=$ change in $K.E.$
$\Rightarrow mg(2R)=\frac{1}{2}m{v}_0^2$
$v_0^2=2g\cdot 2R=4gR$
$v_0=\sqrt{4gR}=2\sqrt{gR}$