A thin homogeneous rod of mass m and length `l` is free to rotate in vertical plane about a horizontal axle pivoted at one end of the rod. A small ballof mass m and charge q is attached to the opposite end of this rod. The whole system is positioned in a horizontal electric field of magnitude `E=(mg)/(2q)`. The rod is released from, shown position from rest.

What is the speed of ball when rod becomes vertical?

Options

A. `sqrt((2gl)/3)`
B. `sqrt((3gl)/4)`
C. `sqrt((3gl)/2)`
D. `sqrt((4gl)/3)`

Answer

C

Explanation

From work energy theorem `1/2 I omega^(2)=mg(l/2)+mgl-qEl`
`1/2 (4/3 ml^(2))omega^(2)=3/2 mgl-(mgl)/2 rArr 2/3 ml^(2)omega^(2)=mgl rArr omega=sqrt((3g)/(2l))`
Speed of ball `=omega l=sqrt((3gl)/2)`