A thick wire is stretched, so that its length become two times. Assuming that there is no change in its density, then what is the ratio of change in resistance of wire to the initial resistance of wire?

Question

Dr. Abubacker · Accepted Answer

(c) Resistance of the wire is given by
$R = \rho\frac{l}{A} = \rho\frac{l^{2}}{Al} = \frac{\rho l^{2}}{V}\ \ \ \ (\because Al = V)$
So,
$R \propto l^{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $(if density remains same)
$or\ \ \ \frac{R^{'}}{R} = \frac{(2l)^{2}}{(l)^{2}} = 4$
$R^{'} = 4R$
Hence, change in resistance
=4R-R=3R
$Therefore,\ \frac{change\ in\ resistance}{original\ resistance} = \frac{3R}{1R} = 3:1$

Q. A thick wire is stretched, so that its length become two times. Assuming that there is no change in its density, then what is the ratio of change in resistance of wire to the initial resistance of wire?

Explanation

Reviewed by Dr. Abubacker

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