Two rods are joined to end, as shown. Both have a cross sectional area of 0.01 cm^2. Each is 1 meter long. One rod is of copper with a resistivity of…

According to the question Length of each $\operatorname{rod} l=1 \mathrm{~m}$rodl=1 m Area of cross-section of cach rod $\mathrm{A}=0.01 \mathrm{~cm}^2=0.01 \times 10^{-4} \mathrm{~m}^2$A=0.01 cm2=0.01×10−4 m2 $\therefore$∴ If $\rho_{c u}=$ρcu​= resistivity of copper and $\rho_{C W}$ρCW​ is resistivity of iron then $$\begin{aligned} & \rho_{c u}=1.7 \times 10^{-6} \Omega \mathrm{cm}=1.7 \times 10^{-8} \Omega \\ & \rho_{f e}=10^{-5} \Omega \mathrm{cm}=10^{-7} \Omega \end{aligned}$$​ρcu​=1.7×10−6Ωcm=1.7×10−8Ωρfe​=10−5Ωcm=10−7Ω​ New Resistance of $\mathrm{Cu}$Cu rod $=\mathrm{R}_{c u}=$=Rcu​= $$\begin{aligned} & \rho_{c u} \frac{1}{\mathrm{~A}} \\ & \text { and Resistance of iron rod } \\ & \rho_{f e}=\rho_{f e} \frac{1}{\mathrm{~A}} \end{aligned}$$​ρcu​ A1​ and Resistance of iron rod ρfe​=ρfe​ A1​​ Both are connected in series then equivalent resistance is $$\begin{aligned} R_{e s}=R_{c e}+R_{F e} & =\rho_{c u} \frac{l}{\mathrm{~A}}+\rho_{f e} \frac{l}{\mathrm{~A}}=\frac{l}{\mathrm{~A}} \\ \left(\rho_{c u}+\rho_{f e}\right) & \\ V & =I R \quad\{\text { where } \mathrm{I}=1 \mathrm{~A}\} \\ & =I\left(R_{c u}+R_{f e}\right) \\ & =1\left(\rho_{c u}+\rho_{f e}\right) \\ & =\left(\rho_{c u}+\rho_{f e}\right)\left(\frac{1}{\mathrm{~A}}\right) \\ & =\left(1.7 \times 10^{-8}+10^{-7}\right) \\ & =10^{-7}(0.17+1)\left(10^6\right) \\ & =1.17 \times 10^{-1} \mathrm{~V} \\ & =0.117 \mathrm{~V} \end{aligned}$$Res​=Rce​+RFe​(ρcu​+ρfe​)V​=ρcu​ Al​+ρfe​ Al​= Al​=IR{ where I=1 A}=I(Rcu​+Rfe​)=1(ρcu​+ρfe​)=(ρcu​+ρfe​)( A1​)=(1.7×10−8+10−7)=10−7(0.17+1)(106)=1.17×10−1 V=0.117 V​