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

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 C
56.25%

$V o l u m e = L \times A = C o n s t a n t = L_{N e w} \times A_{N} e w$

$L_{N e w} = 1 + \frac{25}{100}$ ~=~1.25L\)

$L \times A = 1.25 L \times A_{N e w}$

As R = ρ $\frac{L}{A}$

Final resistance = ${1.25}^{2}$ R = 1.5625R\)

Increase = Final resistance - Initial resistance = 0.5625R

Therefore, Increase in Resistance = 56.25%

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

The correct option is C 56.25% Volume = L×A = Constant = LNew×ANew LNew = 1 + 25100~=~1.25L\) L×A=1.25L×ANew As R = ρLA Final resistance = 1.252R = 1.5625R\) Increase = Final resistance - Initial resistance = 0.5625R Therefore, Increase in Resistance = 56.25%