Q. The resistance of a wire is $R$ ohm. If it is melted and stretched to $n$ times its original length, its new resistance will be
NTA AbhyasNTA Abhyas 2022
Solution:
Resistance is given by,
$R=\frac{\rho l}{A}$ , where $\rho $ is the resistivity, $l$ is the length and $A$ is the area of the cross-section.
$\therefore R=\frac{\rho l^{2}}{A l}=\frac{\rho l^{2}}{V o l u m e}$
As the volume and specific resistance $\rho $ remains constant, so
$R \propto l^{2}...\left(\right.1\left.\right)$
According to question $l_{2}=nl$
$\therefore \frac{R_{2}}{R}=\frac{l_{2}^{2}}{l^{2}}=\frac{n^{2} l^{2}}{l^{2}}$
$\Rightarrow R_{2}=n^{2}R$
$R=\frac{\rho l}{A}$ , where $\rho $ is the resistivity, $l$ is the length and $A$ is the area of the cross-section.
$\therefore R=\frac{\rho l^{2}}{A l}=\frac{\rho l^{2}}{V o l u m e}$
As the volume and specific resistance $\rho $ remains constant, so
$R \propto l^{2}...\left(\right.1\left.\right)$
According to question $l_{2}=nl$
$\therefore \frac{R_{2}}{R}=\frac{l_{2}^{2}}{l^{2}}=\frac{n^{2} l^{2}}{l^{2}}$
$\Rightarrow R_{2}=n^{2}R$