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QJEEAdv15Ph7bac: JEE Advanced 2015 Physics - Class 12 JEEAdv15Ph7bac Solution: In an aluminium (Al) bar of square cross section, a square hole …

In an aluminium (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 ×\times 10 −- 8 Ω\Omega m and 1.0 ×\times 10 −- 7 Ω\Omega m, respectively. The electrical resistance between the two faces P and Q of the composite bar is
JEE Advanced 2015 Paper 1 Offline Physics - Current Electricity Question 20 English

Multiple Choice Options

  • A 247564μΩ{{2475} \over {64}}\mu \Omega
  • B 187564μΩ{{1875} \over {64}}\mu \Omega
  • C 187549μΩ{{1875} \over {49}}\mu \Omega
  • D 2475132μΩ{{2475} \over {132}}\mu \Omega

Step-by-Step Verified Solution

Resistance of a wire, R=ρlAR = {{\rho l} \over A}

For iron (Fe) bar,

ρ\rho = 10 −- 7 Ω\Omega m, l = 50 mm = 50 ×\times 10 −- 3 m

A = (2 mm) ×\times (2 mm) = 4 mm2 = 4 ×\times 10 −- 6 m2

R1=10−7×50×10−34×10−6=1250×10−6{R_1} = {{{{10}^{ - 7}} \times 50 \times {{10}^{ - 3}}} \over {4 \times {{10}^{ - 6}}}} = 1250 \times {10^{ - 6}} Ω\Omega = 1250 μ\mu W

For aluminium (Al) bar,

ρ\rho = 2.7 ×\times 10 −- 8 Ω\Omega m, l = 50 mm = 50 ×\times 10 −- 3 m

A = (72 −- 22) mm2 = 45 mm2 = 45 ×\times 10 −- 6 m2

∴\therefore R2=2.7×10−8×50×10−345×10−6{R_2} = {{2.7 \times {{10}^{ - 8}} \times 50 \times {{10}^{ - 3}}} \over {45 \times {{10}^{ - 6}}}}

=27×5045×10−6=30×10−6= {{27 \times 50} \over {45}} \times {10^{ - 6}} = 30 \times {10^{ - 6}} Ω\Omega = 30 μ\mu Ω\Omega

Potential difference across both bars (resistors) is same so they are in parallel combination.

Equivalent resistance between P and Q is given by

R=R1R2R1+R2=1250×301250+30=125×30128=187564μΩR = {{{R_1}{R_2}} \over {{R_1} + {R_2}}} = {{1250 \times 30} \over {1250 + 30}} = {{125 \times 30} \over {128}} = {{1875} \over {64}}\mu \Omega .

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