A body of mass 1g and carrying a charge 10^–8 C passes from two points P and Q. P and Q are at electric potentials 600V and OV respectively. The velocity of the body at Q is 20 cm s^–1. Its velocity in ms^–1 at P is

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$\sqrt{0 .56}$

$\sqrt{5 .6}$

$\sqrt{0 .056}$

$\sqrt{0 .028}$

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{1}{2} {mv}_{1}^{2} - \frac{1}{2} {mv}_{2}^{2} = (V_{2} - V_{1}) q \\ \frac{1}{2} \times 10^{- 3} (0 .2 \times 0 .2 - v_{2}^{2}) = (600 - 0) \times 10^{- 8} \\ v_{2} = \sqrt{0 .028} \end{array}$

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