In an aluminum (AI) 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 AI and Fe are $2.7 \times 10^{-8} \Omega \mathrm{~m}$ and $1.0 \times 10^{-7} \Omega \mathrm{~m}$, respectively. The electrical resistance between the two faces P and Q of the composite bar is

(B) $\frac{1875}{64} \mu \Omega$

(A) $\frac{2475}{64} \mu \Omega$

(C) $\frac{1875}{49} \mu \Omega$

(D) $\frac{2475}{132} \mu \Omega$

JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Current Electricity

In an aluminum (AI) 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 AI and Fe are

2.7 \times 1 0^{- 8} Ω m

and

1.0 \times 1 0^{- 7} Ω m

, respectively. The electrical resistance between the two faces P and Q of the composite bar is

(A) $\frac{2475}{64} μ Ω$
(B) $\frac{1875}{64} μ Ω$ ✓ Correct
(C) $\frac{1875}{49} μ Ω$
(D) $\frac{2475}{132} μ Ω$

Answer: (B)

Solution

The aluminum and iron resistances are in parallel.

For aluminum:

R_{A l} = \frac{ρ _{A l} \cdot L}{A _{A l}} = \frac{2.7 \times 1 0 ^{- 8} \times 0.05}{( 0.007 ) ^{2} - ( 0.002 ) ^{2}} = \frac{2.7 \times 1 0 ^{- 8} \times 0.05}{45 \times 1 0 ^{- 6}} = 3 \times 1 0^{- 5} Ω

For iron:

R_{F e} = \frac{ρ _{F e} \cdot L}{A _{F e}} = \frac{1.0 \times 1 0 ^{- 7} \times 0.05}{( 0.002 ) ^{2}} = \frac{5 \times 1 0 ^{- 9}}{4 \times 1 0 ^{- 6}} = 1.25 \times 1 0^{- 3} Ω

For parallel combination:

\frac{1}{R _{e q}} = \frac{1}{3 \times 1 0 ^{- 5}} + \frac{1}{1.25 \times 1 0 ^{- 3}} = \frac{1 0 ^{5}}{3} + \frac{0.8 \times 1 0 ^{3}}{1} = \frac{1 0 ^{5} + 2.4 \times 1 0 ^{3}}{3}

\frac{1}{R _{e q}} = \frac{102400}{3} \approx \frac{64}{1875 \times 1 0 ^{- 6}}

R_{e q} = \frac{1875}{64} μ Ω

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