1.

An electron falls freely in electric field of 9.1 xx 10^3 NC^(-1), then acceleration of electron is .......

Answer»

`1.6 xx 10^(13) MS^(-2)`
`1.6 xx 10^(15) cm//s^(2)`
`1.6 xx 10^(15) ms^(-2)`
`1.6 xx 10^(11) ms^(-2)`

Solution :`a = (Ee)/m = (9.1 xx 10^(3) xx 1.6 xx 10^(-19))/(9.1 xx 10^(-31))`
`=1.6 xx 10^(15) m//s^(2)`


Discussion

No Comment Found

Related InterviewSolutions