BITSAT 2021 Paper

Let we have a wire of length $l$, cross-sectional area A, and Resistivity $\rho$Then Resistance of wire is$R=\rho \frac{l}{A}-(1)$The volume of cylindrical wire is the product of its length and cross-sectional area.$V=I \cdot A$If length is increased by 10 percent, the new length l' is$I'=\left|+10\% \,\text{ of }\, I=1.1\right|-(2)$After stretching the cross-sectional area will also change, as volume remains constant. Let the new cross-sectional area be A'$V=I \cdot A = I' A' \cdots \text{(3)}$Putting (2) in (3) we have$I \cdot A = 1.1 I \cdot A'$$A' = \frac{A}{1.1} \cdots (4)$So, the Resistance of this stretched wire is$R' = \rho \frac{l'}{A'} \cdots \text{(5)}$Putting (2) and (4) in (5)$R' = \rho \left( \frac{1.1 l}{\frac{A}{1.1}} \right)$$\Rightarrow R' = \rho \left( \frac{1.1 \times 1.1 l}{A} \right)$$\Rightarrow R' = \rho \left( \frac{1.21 l}{A} \right)$$\Rightarrow R' = 1.21 \left( \rho \frac{l}{A} \right)-(6)$Comparing $\text{Eq}(6)$ and $\text{Eq}(1)$ We get $R' = 1.21 R$So, the resistance increases $1.21$ times.