Current Electricity

JEE Main level, single correct, medium

139 of 380

Question 139 of 380

A Copper (Cu)(Cu) rod of length 25cm25cm and cross-sectional area 3mm23{mm}^{2} is joined with a similar Aluminium (Al)(Al) rod as shown in figure. Find the resistance of the combination between the ends AA and B.B.
(Take resistivity of Copper =1.7×10−8Ωm=1.7\times {10}^{-8}\Omega m, Resistivity of aluminium =2.6×10−8Ωm=2.6\times {10}^{-8}\Omega m)

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Answer: (D) 0.858mΩ0.858m\Omega

R=R1R2R1+R2=ℓA⋅ρ1ρ2ρ1+ρ2R=\dfrac{{R}_{1}{R}_{2}}{{R}_{1}+{R}_{2}}=\dfrac{\ell }{A}\cdot \dfrac{{\rho }_{1}{\rho }_{2}}{{\rho }_{1}+{\rho }_{2}}
R=25×10−23×10−6×1.7×2.6×10−164.3×10−8R=\dfrac{25\times {10}^{-2}}{3\times {10}^{-6}}\times \dfrac{1.7\times 2.6\times {10}^{-16}}{4.3\times {10}^{-8}}
R=0.858mΩR=0.858m\Omega

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