The bob of a pendulum has mass `m= 1kg` and charge `q= 40 muC`. Length of the pendulum is `l=0.9m`. The point of suspension also has the same charge `40 mu C`. What the minimum speed `u` should be imparted to the bob so that it can complete verticel circel ?

Options

A. `6ms^(-1)`
B. `2ms^(-1)`
C. `8ms^(-1)`
D. `4ms^(-1)`

Answer

a

Explanation

Force of repulsion between the charges:
`F=(9xx10^(9)xx40xx40xx10^(-12))/((0.9)^(2))= (160)/(9)N`
Since `Fgt mg`, so string always remains tight at any position even if velocity of bob is zero. So minimum speed at lowest point will be such that it is sufficient to take particle to highest point, i.e., velocity becomes zero at highest point. For this
`u_(min)= sqrt(4gl)=sqrt(4xx10xx0.9)= 6m//s`