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(B) The kinetic energy gained by a charged particle accelerated through a potential difference '$V$' is given by $K = qV$.Since both particles start f

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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 $K = qV$.Since both particles start from rest,the kinetic energy is equal to the work done by the electric field: $\frac{1}{2}mv^2 = qV$.For particle '$A$': $\frac{1}{2}mv_A^2 = qV \implies v_A = \sqrt{\frac{2qV}{m}}$.For particle '$B$': $\frac{1}{2}mv_B^2 = (4q)V \implies v_B = \sqrt{\frac{8qV}{m}}$.Taking the ratio of their speeds $(v_A : v_B)$:$\frac{v_A}{v_B} = \frac{\sqrt{2qV/m}}{\sqrt{8qV/m}} = \sqrt{\frac{2}{8}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.Thus,the ratio of their speeds is $1: 2$.

$A$ particle '$A$' has charge '$+q$' and a particle '$B$' has charge '$+4q$'. Each has the same mass '$m$'. When they are allowed to fall from rest through the same potential difference '$V$',the ratio of their speeds will be (particle $A$ to particle $B$):

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