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Q.

An automobile spring extends 0.2 m for 5000 N load. The ratio of potential energy stored in the spring when it has been compressed by 0.2 m to the potential energy stored in a 10 $μ$ F capacitor at a potential difference of 10,000 V is

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a

2

b

1/2

c

1

d

1/4

answer is C.

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Detailed Solution

For spring, F = k x
$∴ 5000 = k \times 0 .2 or k = 25000 N / m$
Now $\frac{(Potential energy)_{spring}}{(Potential energy)_{capacitor}} = \frac{\frac{1}{2} {kx}^{2}}{\frac{1}{2} {CV}^{2}}$
$= \frac{\frac{1}{2} \times 25000 \times (0 \cdot 2)^{2}}{\frac{1}{2} \times (10 \times 10^{- 6}) \times (10^{4})^{2}} = 1$

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