Physics - Current Electricity Question with Solution | TestHub

Initial radius of wire, $r_1=r$ and final radius of wire, $r_2=r/2=0.5r$. Since volume of the wire after stretch remains constant, therefore $l_1{A}_1=l_2{A}_2$ or $$\frac{l_1}{l_2}=\frac{A_2}{A_1}=\frac{r_2^2}{r_1^2}={\left(\frac{0.5r}{r}\right)}^2=\frac{1}{4}.$$ The resistance $(R)=\rho \frac{l}{A}\propto \frac{l}{A}$. Therefore $\frac{R_1}{R_2}=\frac{l_1}{l_2}\times \frac{A_2}{A_1}=\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$ or $R_2=16R_1=16R$.