Two particle of masses in the ration 1:2 with charges in the ratio 1:1, are placed at rest in a uniform electric field. They are released and allowed to move for the same time. The ratio of their kinetic energies will be finally

Answer

2

Explanation

`a_(1) =(qE)/(m)` and `a_(2)=(qE)/(2m) implies a_(1)=2a_(2)`
`K_(1)=(1)/(2)m_(1)v_(1)^(2)=(1)/(2)m_(1)(a_(1)t)^(2), K_(2)=(1)/(2)m_(2)v_(2)^(2)=1/2m_(2)(a_(2)t)^(2)`
`:. (K_(1))/(K_(2))=(m_(1)a_(1)^(2))/(m_(2)a_(2)^(2))=(m)/(2m)xx(4)/(1)=(2)/(1)`.