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 × 10–8 Ω m and 1.0 × 10–7 Ω m, respectively. The electrical resistance between the two faces P and Q of the composite bar is

Engineering

Physics

Ohm Law and Current Basics

Question

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^–8 Ω m and 1.0 × 10^–7 Ω m, respectively. The electrical resistance between the two faces P and Q of the composite bar is

$\frac{1875}{49} μΩ$

$\frac{2475}{64} μΩ$

$\frac{2475}{132} μΩ$

$\frac{1875}{64} μΩ$

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$R_{Al} = 2 .7 \times 10^{- 8} \times \frac{50}{45 \times 10^{- 3}} = \frac{27 \times 5}{45} \times 10^{- 5} = 30 µ Ω$

$R_{Fe} = 1 .0 \times 10^{- 7} \times \frac{50}{45 \times 10^{- 3}} = \frac{50}{4} \times 10^{- 4} = \frac{5000}{4} µ Ω = 1250 µ Ω$

$R_{g} = \frac{1250 \times 30}{1280} μΩ = \frac{625 \times 3}{64} μΩ = \frac{1875}{64} μΩ$