Physics - Current Electricity Question with Solution | TestHub

According to the question Length of each $\mathrm{rod}l=1\mathrm{m}$ Area of cross-section of cach rod $\mathrm{A}=0.01{\mathrm{cm}}^2=0.01\times 10^{-4}{\mathrm{m}}^2$ $\therefore$ If ${\rho }_{cu}=$ resistivity of copper and ${\rho }_{CW}$ is resistivity of iron then$\begin{array}{rl}& {\rho }_{cu}=1.7\times 10^{-6}\Omega \mathrm{cm}=1.7\times 10^{-8}\Omega \\ & {\rho }_{fe}=10^{-5}\Omega \mathrm{cm}=10^{-7}\Omega \end{array}$ New Resistance of $\mathrm{Cu}$ rod $={\mathrm{R}}_{cu}=$ $\begin{array}{rl}& {\rho }_{cu}\frac{1}{\mathrm{A}}\\ & \text{ and Resistance of iron rod }\\ & {\rho }_{fe}={\rho }_{fe}\frac{1}{\mathrm{A}}\end{array}$ Both are connected in series then equivalent resistance is $\begin{array}{rl}{R}_{es}=R_{ce}+R_{Fe}& ={\rho }_{cu}\frac{l}{\mathrm{A}}+{\rho }_{fe}\frac{l}{\mathrm{A}}=\frac{l}{\mathrm{A}}\\ \left({\rho }_{cu}+{\rho }_{fe}\right)& \\ V& =IR\{\text{ where }\mathrm{I}=1\mathrm{A}\}\\ & =I\left(R_{cu}+R_{fe}\right)\\ & =1\left({\rho }_{cu}+{\rho }_{fe}\right)\\ & =\left({\rho }_{cu}+{\rho }_{fe}\right)\left(\frac{1}{\mathrm{A}}\right)\\ & =\left(1.7\times 10^{-8}+10^{-7}\right)\\ & =10^{-7}(0.17+1)\left(10^6\right)\\ & =1.17\times 10^{-1}\mathrm{V}\\ & =0.117\mathrm{V}\end{array}$