Question

A wire is stretched so as to change its diameter by $0.25\%$. The percentage change in resistance is

• A. $4.0\%$
• B. $2.0\%$
• C. $1.0\%$
• D. $0.5\%$

Answer 1

Answer: (C)

On stretching, volume $(V)$ remains constant.
So, $V=Al$ or $l=V/A$
Now, $R=\frac{\rho l}{A}=\frac{\rho V}{A^2}=\frac{\rho V}{{\pi }^2{D}^4/16}=\frac{16\rho V}{{\pi }^2{D}^4}$
Taking logarithm of both the side and differentiating it, we get
$\frac{\Delta R}{R}=-4\frac{\Delta D}{D}\text{ or }\frac{\Delta R}{R}=-4\times (-0.25)=1.0\%$
Resistance will increase.