Physics - Current Electricity Question with Solution | TestHub

Given$l^{\prime }=l+100\% of l=2l$

Initial volume = final volume

ie,$\pi r^{2}l=\pi r^{\prime 2}{l}^{\prime }$

$⟹ r^{\prime 2}=\frac{r^{2}l}{l^{\prime }}=r^2\times \frac{l}{2l}$

$⟹ r^{\prime 2}=\frac{r^2}{2}$

$\therefore R^{\prime }=\rho \frac{l^{\prime }}{A^{\prime }}=\rho \frac{2l}{\pi r^{\prime 2}} \left(\because R=\frac{\rho l}{A}\right)$

$=\frac{\rho 4l}{\pi r^2}$= 4R
=
Thus,$∆R=R^{\prime }-R=4R-R=3R$

$\therefore \%∆R=\frac{3R}{R}\times 100\%$

$=300\%$