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A mass of \( 1 \mathrm{Kg} \) carrying a charge of \( 2 \mathrm{C} \) is accelerated through a potential of \( 1 \mathrm{~V} \). The velocity
acquired by it is
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acquired by it is
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Verified Answer
The correct answer is:
\( 2 m s^{-1} \)
Given, a mass is accelerated through a potential. Therefore,
$\frac{1}{2} m v^{2}=q V$
Here, $\mathrm{m}=1 \mathrm{~kg}$; $\mathrm{q}=2 \mathrm{C} ; \mathrm{V}=1$
$\therefore \frac{1}{2} \times 1 \times V^{2}=2 \times 1$
$\Rightarrow V^{2}=4 \Rightarrow V=2$
Thus, velocity acquired by mass is $2 \mathrm{~ms}^{-1}$
$\frac{1}{2} m v^{2}=q V$
Here, $\mathrm{m}=1 \mathrm{~kg}$; $\mathrm{q}=2 \mathrm{C} ; \mathrm{V}=1$
$\therefore \frac{1}{2} \times 1 \times V^{2}=2 \times 1$
$\Rightarrow V^{2}=4 \Rightarrow V=2$
Thus, velocity acquired by mass is $2 \mathrm{~ms}^{-1}$
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