A particle A has charge +q and a particle B h | vedclass

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(B) The kinetic energy gained by a charged particle accelerated through a potential difference $V$ is given by $K = QV = \frac{1}{2}mv^2$.From this,th

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$1:2$

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(B) The kinetic energy gained by a charged particle accelerated through a potential difference $V$ is given by $K = QV = \frac{1}{2}mv^2$.From this,the speed $v$ is given by $v = \sqrt{\frac{2QV}{m}}$.Since both particles have the same mass $m$ and fall through the same potential difference $V$,the speed is proportional to the square root of the charge: $v \propto \sqrt{Q}$.Therefore,the ratio of their speeds is $\frac{v_A}{v_B} = \sqrt{\frac{Q_A}{Q_B}}$.Substituting the given values $Q_A = q$ and $Q_B = 4q$,we get $\frac{v_A}{v_B} = \sqrt{\frac{q}{4q}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.Thus,the ratio is $1:2$.

$A$ particle $A$ has charge $+q$ and a particle $B$ has charge $+4q$,with each of them having the same mass $m$. When allowed to fall from rest through the same electric potential difference,the ratio of their speeds $\frac{v_A}{v_B}$ will be:

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