Question

Find the percentage increase in resistance of a wire when it is stretched uniformly so that its length increases by $ 0.5\% $

• A. $ 1 \% $
• B. $ 2 \% $
• C. $ 2.5 \% $
• D. $ 3 \% $

Answer 1

Answer: (A)

As volume remains constant in the process of stretching
$\therefore V=Al$ = constant
$0=\frac{\Delta\,A}{A}+\frac{\Delta\,l}{l}$ or
$\left|\frac{\Delta A}{A}\right|=\left|\frac{\Delta l}{l}\right| \ldots\left(i\right) $
Resistance, $R=\rho \frac{l}{A}$
$\therefore \frac{\Delta R}{R}=\left|\frac{\Delta l}{l}\right|+\left|\frac{\Delta A}{A}\right|$ $(\because \rho$ is constant)
$\frac{\Delta R}{R}=2\left|\frac{\Delta l}{l}\right|$ (using eqn. (i))
$\therefore \%$ increase in resistance of wire
$=2\times0.5\% $
$=1\%$