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Question

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


A

25%

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B

50%

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C

56.25%

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D

77.33%

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Solution

The correct option is C

56.25%


We know that, resistance R = ρLA and volume = L x A
where,
L = length of the wire before stretching and
A = area of the wire before stretching
L' = length of the wire after stretching and
A' = area of the wire after stretching
Given:
resistance of wire R=1 Ω
increase in length by percentage = 25%
new length, L = (L + 25% of L)
or L=1.25×L
Since the length of the wire increased and the total volume is constant
L×A = L×A
1×A = 1.25×A
or A=0.8A
New resistance, R=ρ1.25L0.8A
or, R=1.5625R
Increase in resistance = R' - R
Since R=1 Ω
Increase in resistance = 1.5625 - 1 = 0.5625
% increase in resistance = RRR×100
= 0.56251×100=56.25
Therefore, increase in resistance = 56.25%.


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