Physics - Current Electricity Question with Solution | TestHub

Let the length of the wire bel, radius of the wire be r

$\therefore$Resistance$R=\rho \frac{l}{\pi r^2}\rho =$resistivity of the wire

Now$l$is increased by 100%

$\therefore l^{\text{'}}=l+\frac{100}{100}l=2l$

As length is increased, its radius is going to be decreased in such a way that the volume of the cylinder remains constant.

$\pi r^2\times l=\pi {r^{\text{'}}}^2\times l^{\text{'}}\Rightarrow r^{\text{'}2}=\frac{r^2\times l}{l^{\text{'}}}=\frac{r^2\times l}{2l}=\frac{r^2}{2}$

$\therefore$The new resistance$R^{\text{'}2}=\rho \frac{l^{\text{'}}}{\pi r^{\text{'}2}}=\rho \frac{2l}{\pi \times \frac{r^2}{2}}=4R$

$\therefore$Change in resistance${=R}^{\text{'}}-R=3R$

$\therefore$% change$=\frac{3R}{R}\times 100\%=300\mathrm{\%.}$