A ball of mass 1 g having a charge of 20 `mu C ` is tied to one end of a string of length 0.9 m can rotate in a vertically plane in a uniform electric minimum horizontal velocity that must be given to the ball at the lowest position so that is complete the verticle circle is `(let g= 10 ms^(-1))`

Options

A. `9 ms^(-1)`
B. `18 ms^(-1)`
C. `36 ms^(-1)`
D. `6 ms^(-1)`

Answer

D

Explanation

Given mass of ball m =1 g`=(1)/(10^(3)) kg`
charge on the ball `q= 20 mu C = 20 xx 10^(6) c`
length of the string centre of one end r=0.9 m
electric field E = `100 N//c`
and acceleration due to
gravity g= `10 m// s^(2)`
now according to the question

Force exerted by the electric field due to charge on ball is given as
Force = charge on the ball `xx` Electric field
net effective force on the ball
`F_(off ) = F_(2 - F_(1)`
from Eqs . (ii) and (i ) we get
`F_(off) = 10 xx 10^(-3) - 2xx 10^(-3) rArr f_(off) = 8 xx 10^(-3) N`
Now effective gravitational acceleration at lowest postion in given as
`F_(off) = mg_(off )`
`g_(eff ) = (F_(eff))/(m ) = (8 xx 10^(-3))/(10^(-3)) = 8m//s^(2)`
The minimum horizontal velocity that mut be given to the ball at the lowest position is
`v= sqrt(5g_(eff)) rArr v = sqrt(5xx8 xx.(9)/(10))`
`v= sqrt(4xx9) = sqrt(36) rArr v= 6 m//s`