Question

If radius of a resistor is decreased by $0.5\%$ by stretching then percentage change in resistance

• A. $0.5\%$
• B. $1\%$
• C. $1.5\%$
• D. $2\%$

Answer 1

The correct option is D $2\%$

Resistance of the resistor is given by,
$R=\frac{\rho l}{A}=\frac{\rho l\times A}{A\times A}=\frac{Mass}{A^2}=\frac{Mass}{\pi r^4}$
$\Rightarrow R\propto \frac{1}{r^4}$

Differentiating, we get, $dR\propto -\frac{4}{r^5}dr$
$\therefore \frac{dR}{R}=\frac{-\frac{4}{r^5}}{\frac{1}{r^4}}dr=-\frac{4dr}{r}$

Hence, percent change in resistance is given by
$\frac{dR}{R}\times 100=-4\times \frac{dr}{r}\times 100$
Where $\frac{dR}{R}$ is change in resistance and $\frac{dr}{r}$ is change in radius = -0.5% (here, decrease in radius is indicated by negative sign).
$\frac{dR}{R}\times 100=-4\times (-0.5)$
$\frac{dR}{R}\times 100=2\%$