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PhysicsMultiple correct+4 / -2Question 17 of 60

JEE Advanced 2015 · Paper 1, Physics question 17

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 Ω2.7\times10^{-8}\ \Omega m and 1.0×10−7 Ω1.0\times10^{-7}\ \Omega m, respectively. The electrical resistance between the two faces P and Q of the composite bar is

A bar of square cross-section 7 mm × 7 mm and length 50 mm made of Al, with a central square hole 2 mm × 2 mm running the full length, filled with Fe. P is the near end face and Q the far end face.
Choose one or more options
Answer and worked solution

Official answer: B

Two conductors in parallel along the 50 mm length.

  • Al: area 49−4=4549-4=45 mm², so RAl=2.7×10−8×0.0545×10−6=30 μΩR_{Al}=\dfrac{2.7\times10^{-8}\times0.05}{45\times10^{-6}}=30\ \mu\Omega.
  • Fe: area 4 mm², so RFe=10−7×0.054×10−6=1250 μΩR_{Fe}=\dfrac{10^{-7}\times0.05}{4\times10^{-6}}=1250\ \mu\Omega.

Combine.
R=30×12501280=187564 μΩ.R=\frac{30\times1250}{1280}=\frac{1875}{64}\ \mu\Omega.

(B)

Topics:Ohm's law and resistivitySeries and parallel resistors
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