A body of mass $1 g$ and carrying a charge $10^{- 8} C$ passes through two points $P$ and $Q$ . $P$ and $Q$ are at electric potential $600 V$ and $0 V$ respectively. The velocity of the body at $Q$ is $20 \times 10^{- 2} {ms}^{- 1}$ . Its velocity at $P$ is $\sqrt{x \times 10^{- 3}} {ms}^{- 1}$ . The value of $x$ is

28.00

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28.0

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Solution

Given:
$m = 1 g = 10^{- 3} kg q = 10^{- 8} C V_{P} = 600 V V_{Q} = 0 V v_{Q} = 20 \times 10^{- 2} {ms}^{- 1}$

From work-energy theorem,

Work done $=$ Change in KE

$q (V_{P} - V_{Q}) = \frac{1}{2} m (v_{Q}^{2} - v_{P}^{2})$

$\Rightarrow 10^{- 8} (600 - 0) = \frac{1}{2} \times 10^{- 3} \times (v_{Q}^{2} - v_{P}^{2})$

$\Rightarrow 0.012 = (20 \times 10^{- 2})^{2} - v_{P}^{2}$

$\Rightarrow v_{P}^{2} = 0.04 - 0.012 = 0.028$

$\Rightarrow v_{P} = \sqrt{0.028} {ms}^{- 1} = \sqrt{28 \times 10^{- 3}} {ms}^{- 1}$

$∴ x = 28$

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