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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)
  1. 344.43

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Solution

The correct option is A 344.43
Given:
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%(10025)%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
At2
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=R2R1=133.3330
=103.33 Ω

Percentage change in the resistance,
ΔRR1×100=103.3330×100
=344.43%

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