In an aluminum (Al) bar of square cross section, a square hole is drilled and is filled with iron (Fe) as shown in the figure. The electrical resistivities of Al and Fe are 2.7×10−8Ω m and 1.0×10−7Ω m, respectively. The electrical resistance between the two faces P and Q of the composite bar is
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a
247564μΩ
b
187564μΩ
c
187549μΩ
d
2475132μΩ
answer is B.
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Detailed Solution
Both rods are in parallel, as they will be connected to same potential difference.1R=1RAl+1RFe⇒1R=AAlρAl+AFeρFe1ℓ⇒1R=72−222.7+221010−610−8×150×10−3⇒1R=452.7+252×103=45027+252×103⇒1R=503+252×103=25615×2×103⇒R=15512×103μΩ=187564μΩ