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, $V=A l$
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 \%$