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

The correct option is C 1.0%

On stretching volume (V) remains constant.
So, V = Al
Or $l=\frac{V}{A}$
$\therefore R=\rho .\frac{l}{A}=\rho .\frac{l}{A}=\frac{\rho V}{{\pi }^2\frac{D^4}{16}}=\frac{16\rho V}{{\pi }^2{D}^4}$
Taking logarithm of both sides and differentiating, we get
$\frac{\Delta R}{R}=-4\frac{\Delta D}{D}Or\frac{\Delta R}{R}=4\times 0.25=1.0\%$