TS EAMCET 18-July-2022 Shift 2 Solved Paper

A wire is stretched such that its resistance increases by $6 \%$ .If we assume initial resistance to be $100 \Omega$ Then the new resistance will be given by $R_{\;\text{new}\;}=100 \Omega+\;\frac{6}{100}\times 100 \Omega=106 \Omega$We know that, the resistance is given by $R=\rho \;\frac{I}{A} \;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot(i)$Where, aréa $(A)$ can be expressed in terms of volume $(V)$ is $A=\;\frac{V}{l}$Equation (i) then gives us, $R=\rho \;\frac{l^{2}}{V}$Since, the resistivity and volume are going to remain the same, only change in length will be observed.So, $\;\; l^{2}=\;\frac{R V}{\rho} \;\; \ldots$..... (ii) (Original length)Let new length be $l_{\;\text{new}\;}$ , so $l_{\;\text{new}\;}^{2}=\;\frac{R_{\;\text{new}\;} V}{\rho}\;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot(iii)$Dividing Eq. (iii) by Eq. (ii), we get $\;\frac{l_{\;\text{new}\;}^{2}}{l^{2}}=\;\frac{R_{\;\text{new}\;}}{R}=\;\frac{106}{100} \;\text{". "}\;$ $\Rightarrow \;\; l_{\;\text{new}\;}=1.029 l$ $\%$ change in length $\;= \left(\frac{l_{\;\text{new}\;}-l}{l}\right)\times 100 \%= \left(\frac{1.029 l-l}{l}\right)\times 100$ $\;=0.029\times 100=29 \%$ $\;\; \approx 3 \%$