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:
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 = R′−RR×100
= 0.56251×100=56.25
Therefore, increase in resistance = 56.25%.