JEE MAIN - Physics (2022 - 26th June Morning Shift - No. 12)
An aluminium wire is stretched to make its length, 0.4% larger. The percentage change in resistance is :
0.4%
0.2%
0.8%
0.6%
Explanation
$$R = {{\rho l} \over A}$$
Also volume will remain constant
i.e., Al = constant $$ \Rightarrow A \propto {1 \over l}$$
$$\therefore$$ $$R \propto {l^2}$$
$${{\Delta R} \over R} = 2{{\Delta l} \over l} = 0.8$$
Comments (0)
