Question

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

• A. $4\%$
• B. $2\%$
• C. $1\%$
• D. $0.5\%$

Answer 1

The correct option is C $1\%$

$R=\frac{pl}{A}=\frac{PV}{A^2}=\frac{PV}{{\pi }^2{r}^4}R\alpha r^{-4}\%\frac{\Delta R}{R}=-4\frac{\Delta r}{r}\%=-4\left[0.25\%\right]=-1\%$