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IV. Numerical Problems · Q5

Q.A rod is made up of two different materials, joined end to end, of lengths 25 cm and 70 cm respectively. Both have square cross sections of 3 mm side. The resistivity of the first material is 4×10−3 Ωm4\times10^{-3}\ \Omega\text{m} and that of the second material is 5×10−3 Ωm5\times10^{-3}\ \Omega\text{m}. What is the resistance of the rod between its ends?

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Step 1. Cross-sectional area: side = 3 mm = 3×10−33\times10^{-3} m, so A=(3×10−3)2=9×10−6 m2A=(3\times10^{-3})^2=9\times10^{-6}\ \text{m}^2 (same for both segments).

Step 2. First segment: length l1=25l_1=25 cm =0.25=0.25 m, resistivity ρ1=4×10−3 Ωm\rho_1=4\times10^{-3}\ \Omega\text{m}. R1=ρ1l1/A=(4×10−3)(0.25)/(9×10−6)=(1×10−3)/(9×10−6)≈111.1 ΩR_1=\rho_1 l_1/A = (4\times10^{-3})(0.25)/(9\times10^{-6}) = (1\times10^{-3})/(9\times10^{-6}) \approx 111.1\ \Omega.

Step 3. Second segment: length l2=70l_2=70 cm =0.70=0.70 m, resistivity ρ2=5×10−3 Ωm\rho_2=5\times10^{-3}\ \Omega\text{m}. R2=ρ2l2/A=(5×10−3)(0.70)/(9×10−6)=(3.5×10−3)/(9×10−6)≈388.9 ΩR_2=\rho_2 l_2/A = (5\times10^{-3})(0.70)/(9\times10^{-6}) = (3.5\times10^{-3})/(9\times10^{-6}) \approx 388.9\ \Omega. …

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