A wire 1 m long has a resistance of 1Ω. If it is uniformly stretched, so that its length increases by 25%, then its resistance will increase by:

25%

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50%

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56.25%

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77.33%

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Solution

The correct option is D. 56.25%.
Given,
Initial resistance, $R = \frac{ρ l}{A}$

When length is increased by 25%, $l^{'} = 1.25 l$

So, new cross-sectional area, $A^{'} = \frac{A}{1.25}$ (Since volume = Al = constant)

Final resistance $= \frac{ρ l^{'}}{A^{'}} = (1.25)^{2} \times \frac{ρ l}{A} = (1.25)^{2} R = 1.5625 R$

Therefore, percentage increase in resistance = 56.25%.

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A wire 1 m long has a resistance of 1Ω. If it is uniformly stretched, so that its length increases by 25%, then its resistance will increase by:

The correct option is D. 56.25%.Given,Initial resistance, R=ρlA When length is increased by 25%, l′=1.25l So, new cross-sectional area, A′=A1.25 (Since volume = Al = constant) Final resistance =ρl′A′=(1.25)2×ρlA=(1.25)2R=1.5625R Therefore, percentage increase in resistance = 56.25%.