A wire of a certain material is stretched slowly by ten per cent. Its new resistance and specific resistance become respectively

Question

Aicharya · Accepted Answer

Key Idea : In stretching, specific resistance remains unchanged.
After stretching, specific resistance $(\rho)$ will remain same.
Original resistance of the wire,
$R=\frac{\rho l}{A}$
$R \propto \frac{l}{A} \text { or } R \propto \frac{l^2}{V}$ (as $V=A l$)
and $R^{\prime} \propto \frac{(l+10 \% l)^2}{V}$
Therefore, $\frac{R^{\prime}}{R}=\frac{\left(l+\frac{10}{100} l\right)^2}{l^2}$
$\frac{R^{\prime}}{R}=\frac{\left(\frac{11 l}{10}\right)^2}{l^2}=\frac{121}{100}$
$R^{\prime}=1.21 R$

A wire of a certain material is stretched slowly by ten per cent. Its new resistance and specific resistance…

Solution