Question

A particle of mass m and charge q is attached to a string of length $l$. It is whirled into a vertical circle in field E as shown. What is the speed given to the particles at point B so that tension in the string when the particle is at A is ten times the weight of particle?

• A. $\sqrt{5gl}$
• B. ${\left(\frac{qEl}{m}\right)}^{1/2}$
• C. $\sqrt{5(g-\frac{qE}{m})l}$
• D. $\sqrt{5(g+\frac{qE}{m})l}$

Answer 1

Answer: (D)

$\sqrt{5(g+\frac{qE}{m})l}$

Energy Conservation between A & B:

$\frac{1}{2}m{v}_B^2+mg\left(2l\right)=\frac{1}{2}m{v}_A^2+qE\Delta d\Rightarrow m{v}_A^2=m{v}_B^2+4mgl-4qEl$

Applying given condition at A: tension in the string when the particle is at A is ten times the weight of particle

$\frac{m{v}_A^2}{l}+mg-qE=10mg\Rightarrow \frac{m{v}_B^2+4mgl-4qEl}{l}+mg-qE=10mg\Rightarrow \frac{m{v}_B^2}{l}+5mg-5qE=10mg\Rightarrow v_B=\sqrt{5(g+\frac{qE}{m})l}$