A wire is stretched so as …
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Gaurav Seth 4 years, 9 months ago
Let
X, be the initial length, and r
Y be the cross-sectional area
R be the resistance of wire.
Let X', Y' and R' be the new length, cross- sectional area and resistance of wire after the length of wire is increased.
Length is increased by 0.1%, therefore,
L' = L+(0.1/100)L
L' = 1.001 L
Since the volume of wire remains same, thus,
V' = V
A'*L' = A*L
A' * 1.001 L = A*L
A' = A/1.001
New Resistance,
R' = ρL'/A'
= ρ*1.001 L/(A/1.001)
= (1.001)²(ρL/A)
R' = 1.002*R
% change in resistance = (Final resistance - Initial resistance)/Initial resistance * 100%
= (R' - R)/R*100%
= (1.002R - R)/R*100%
= 0.002 * 100%
= 0.2%
Thus, resistance of wire is increased by 0.2%.
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