Two particles of masses in the ratio 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 :

2 : 1

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8 : 1

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4 : 1

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1 : 4

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Solution

The correct option is A 2 : 1
As charge on both particles is same, so electric force acting on them will be same. Since the particles are allowed to move for the same time, their final momentum will be same.
Because change in momentum $=$ impulse $=$ force $\times$ time, so
$K E = \frac{P^{2}}{2 m}$ gives $K E \propto \frac{1}{m}$
$\frac{K E_{1}}{K E_{2}} = \frac{m_{2}}{m_{1}} = 2 : 1$ .

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