A capacitor of capacity $10 μ F$ is charged to a potential of $10000 V$ and a wire is stretched by $0.2 m$ by a force of $5000 N$ . The ration of the potential energies stored in them will be

1

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500

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0.002

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0.0001

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Solution

The correct option is A $1$
The potential energy stored in a capacitor
$V_{1} = \frac{1}{2} C V^{2}$
$= \frac{1}{2} \times 10 \times 10^{- 6} \times {(10^{4})}^{2} . . . . . . . . . . . . (1)$
Work done in stretching the wire is
$W = P E = a v e r a g e f o r c e \times Δ L$
$V_{2} = \frac{F}{2} Δ L$
$= \frac{5000}{2} \times 0.2........... (2)$

Dividing (1) by (2)
$\frac{V_{1}}{V_{2}} = \frac{0.5 \times 10^{3}}{500}$
$= 1$
$∴ V_{1} : V_{2} = 1 : 1$

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