Question

# A 10 cm wire of resistance 30 Ω is stretched to 25 cm, if the thickness of the wire is reduced by 25% in the process, what is the approximate percentage change in the resistance? (Include the sign)344.43

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Solution

## The correct option is A 344.43Given: Initial length of the wire, l1=10 cm Final length of the wire, l2=25 cm Initial resistance of the wire, R1=30 Ω Let the final resistance of the wire be R. If the initial thickness (diameter) of the wire is t, then the final thickness is reduced by 25%⟹(100−25)%t=75%t =0.75t Since it is the same wire, the resistivity (ρ) is going to be the same before and after the stretching. We know that ρ=RAl ⟹R1A1l1=R2A2l2 ---- (1) Area of the wire, A=π×(t2)2 A∝t2 From (1): R1t21l1=R2t22l2 ⟹R2=R1l2l1×t21t22 ⟹R2=30×2510×t2(0.75t)2 ⟹R2=75×1(0.75)2=4003 ⟹R2=133.33 Ω Change in the resistance, ΔR=R2−R1=133.33−30 =103.33 Ω Percentage change in the resistance, ΔRR1×100=103.3330×100 =344.43%

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