Question

A wire is stretched 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 $l=\frac{V}{A}$
$\because R=\rho \cdot \frac{l}{A}$
$=\rho \cdot \frac{V}{A^2}=\frac{\rho V}{{\pi }^2{D}^4/16}$
$=\frac{16\rho V}{{\pi }^2{D}^4}$
Taking logarithm of both sides and differentiating, we get
$\frac{\Delta R}{R}=-4\frac{\Delta D}{D}$
or $\frac{\Delta R}{R}=-4\times 0.25=1.0\%$