Two charged particles of masses m and 2m have charges +2q and +q respectively. They are kept in a uniform electric field far away from each other and then allowed to move for the same time. Find the ratio of their kinetic energies.

8:1

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1:4

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Solution

The correct option is D 8:1
Let electronic field= $E$
$F_{A} = 2 E q, F_{B} = E q$
$G_{A} = \frac{2 E q}{m}, a_{B} = \frac{E q}{2 m}$
Let time be $t$
$v_{A} = \frac{2 E q t}{m}, v_{B} = \frac{E q t}{2 m} \frac{{K E}_{A}}{K E_{B}} = {(\frac{2 E q t}{m})}^{2} m / {(\frac{E q t}{2 m})}^{2} 2 m = \frac{4 / 1}{2 / 4} = 8 / 1 \Rightarrow 8 : 1$

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Two charged particles of masses m and 2m have charges +2q and +q respectively. They are kept in a uniform electric field far away from each other and then allowed to move for the same time. Find the ratio of their kinetic energies.

The correct option is D 8:1Let electronic field=EFA=2Eq,FB=EqGA=2Eqm,aB=Eq2mLet time be tvA=2Eqtm,vB=Eqt2mKEAKEB=(2Eqtm)2m/(Eqt2m)22m=4/12/4=8/1⇒8:1