Question

A mass $m$ has given a charge $+q$ is attached at the end of a light rod. A horizontal uniform electric field is exist as shown in the figure. If the mass is released from the position shown in figure, find the angular velocity of the rod when it passes through the bottom most position.

• A. $\sqrt{\frac{3g}{l}}$
• B. $\sqrt{\frac{g}{l}}$
• C. $\sqrt{\frac{2g}{l}}$
• D. $\sqrt{\frac{5g}{l}}$

Answer 1

The correct option is B $\sqrt{\frac{g}{l}}$

According to law of consrvation of energy

$mgl[1-\cos \theta ]=\frac{1}{2}m{v}^2-\left(\Delta v\right)q$

$mgl[1-\cos \theta ]=\frac{1}{2}m[l\omega {]}^2-E(l\sin \theta )q$


$mgl[1-\cos {45}^{\circ }]=\frac{1}{2}m{l}^2{\omega }^2-\frac{mg}{q}l\sin \theta q$

$g[1-\frac{1}{\sqrt{2}}]=\frac{1}{2}l{\omega }^2-g\sin {45}^{\circ }$

$g-\frac{g}{\sqrt{2}}+\frac{g}{\sqrt{2}}=\frac{1}{2}l{\omega }^2$

$\omega =\sqrt{\frac{2g}{l}}$