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Current Electricity question

2015 · Shift 1 · Q56
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Current Electricity question

2015 · Shift 1 · Q56

JEE AdvancedPhysicsCurrent ElectricityMultiple correct+4 / −2
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\OmegaΩ m and 1.0 ×\times× 10 −-− 7 Ω\OmegaΩ 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 21 English
  1. A
    247564μΩ{{2475} \over {64}}\mu \Omega642475​μΩ
  2. B
    187564μΩ{{1875} \over {64}}\mu \Omega641875​μΩ
  3. C
    187549μΩ{{1875} \over {49}}\mu \Omega491875​μΩ
  4. D
    2475132μΩ{{2475} \over {132}}\mu \Omega1322475​μΩ
View written solutionFree

Correct answer: B

Step-by-step Derivation:

  1. Identify the circuit configuration: The composite bar consists of an Aluminium (Al) part and an Iron (Fe) part. Since the current flows from face P to face Q, the potential difference across both the Al and Fe sections is the same. Therefore, the Al and Fe parts act as two resistors connected in parallel.

  2. Formula for Equivalent Resistance: The equivalent resistance ReqR_{eq}Req​ of two resistors RAlR_{Al}RAl​ and RFeR_{Fe}RFe​ in parallel is given by: 1Req=1RAl+1RFe  ⟹  Req=RAlRFeRAl+RFe{1 \over R_{eq}} = {1 \over R_{Al}} + {1 \over R_{Fe}} \implies R_{eq} = {{R_{Al} R_{Fe}} \over {R_{Al} + R_{Fe}}}Req​1​=RAl​1​+RFe​1​⟹Req​=RAl​+RFe​RAl​RFe​​ The resistance R of a conductor is given by the formula: R=ρLAR = \rho {L \over A}R=ρAL​ where ρ\rhoρ is the resistivity, L is the length, and A is the cross-sectional area.

  3. List Given Values and Convert to SI units:

    • Length of the bar, L=50 mm=50×10−3 mL = 50 \text{ mm} = 50 \times 10^{-3} \text{ m}L=50 mm=50×10−3 m
    • Resistivity of Al, ρAl=2.7×10−8 Ωm\rho_{Al} = 2.7 \times 10^{-8} \, \Omega \text{m}ρAl​=2.7×10−8Ωm
    • Resistivity of Fe, ρFe=1.0×10−7 Ωm\rho_{Fe} = 1.0 \times 10^{-7} \, \Omega \text{m}ρFe​=1.0×10−7Ωm
    • Outer side of Al bar, aout=7 mm=7×10−3 ma_{out} = 7 \text{ mm} = 7 \times 10^{-3} \text{ m}aout​=7 mm=7×10−3 m
    • Side of Fe bar (inner hole), ain=2 mm=2×10−3 ma_{in} = 2 \text{ mm} = 2 \times 10^{-3} \text{ m}ain​=2 mm=2×10−3 m

  4. Calculate the Resistance of the Aluminium Part (RAlR_{Al}RAl​):

    • First, find the cross-sectional area of the Al part, AAlA_{Al}AAl​. AAl=aout2−ain2=(7×10−3)2−(2×10−3)2A_{Al} = a_{out}^2 - a_{in}^2 = (7 \times 10^{-3})^2 - (2 \times 10^{-3})^2AAl​=aout2​−ain2​=(7×10−3)2−(2×10−3)2 AAl=(49×10−6)−(4×10−6)=45×10−6 m2A_{Al} = (49 \times 10^{-6}) - (4 \times 10^{-6}) = 45 \times 10^{-6} \text{ m}^2AAl​=(49×10−6)−(4×10−6)=45×10−6 m2
    • Now, calculate the resistance RAlR_{Al}RAl​. RAl=ρAlLAAl=(2.7×10−8)×50×10−345×10−6R_{Al} = \rho_{Al} {L \over A_{Al}} = (2.7 \times 10^{-8}) \times {{50 \times 10^{-3}} \over {45 \times 10^{-6}}}RAl​=ρAl​AAl​L​=(2.7×10−8)×45×10−650×10−3​ RAl=2.7×5045×10−8−3+6=13545×10−5=3×10−5 ΩR_{Al} = {{2.7 \times 50} \over {45}} \times 10^{-8-3+6} = {{135} \over {45}} \times 10^{-5} = 3 \times 10^{-5} \, \OmegaRAl​=452.7×50​×10−8−3+6=45135​×10−5=3×10−5Ω

  5. Calculate the Resistance of the Iron Part (RFeR_{Fe}RFe​):

    • First, find the cross-sectional area of the Fe part, AFeA_{Fe}AFe​. AFe=ain2=(2×10−3)2=4×10−6 m2A_{Fe} = a_{in}^2 = (2 \times 10^{-3})^2 = 4 \times 10^{-6} \text{ m}^2AFe​=ain2​=(2×10−3)2=4×10−6 m2
    • Now, calculate the resistance RFeR_{Fe}RFe​. RFe=ρFeLAFe=(1.0×10−7)×50×10−34×10−6R_{Fe} = \rho_{Fe} {L \over A_{Fe}} = (1.0 \times 10^{-7}) \times {{50 \times 10^{-3}} \over {4 \times 10^{-6}}}RFe​=ρFe​AFe​L​=(1.0×10−7)×4×10−650×10−3​ RFe=504×10−7−3+6=12.5×10−4=1.25×10−3 ΩR_{Fe} = {{50} \over {4}} \times 10^{-7-3+6} = 12.5 \times 10^{-4} = 1.25 \times 10^{-3} \, \OmegaRFe​=450​×10−7−3+6=12.5×10−4=1.25×10−3Ω

  6. Calculate the Equivalent Resistance (ReqR_{eq}Req​):

    • Using the parallel combination formula: Req=RAlRFeRAl+RFeR_{eq} = {{R_{Al} R_{Fe}} \over {R_{Al} + R_{Fe}}}Req​=RAl​+RFe​RAl​RFe​​
    • Let's express both resistances with the same power of 10 for easier addition. RAl=3×10−5Ω=0.3×10−4ΩR_{Al} = 3 \times 10^{-5} \Omega = 0.3 \times 10^{-4} \OmegaRAl​=3×10−5Ω=0.3×10−4Ω RFe=12.5×10−4ΩR_{Fe} = 12.5 \times 10^{-4} \OmegaRFe​=12.5×10−4Ω
    • Sum of resistances: RAl+RFe=(0.3+12.5)×10−4=12.8×10−4 ΩR_{Al} + R_{Fe} = (0.3 + 12.5) \times 10^{-4} = 12.8 \times 10^{-4} \, \OmegaRAl​+RFe​=(0.3+12.5)×10−4=12.8×10−4Ω
    • Product of resistances: RAlRFe=(0.3×10−4)×(12.5×10−4)=3.75×10−8 Ω2R_{Al} R_{Fe} = (0.3 \times 10^{-4}) \times (12.5 \times 10^{-4}) = 3.75 \times 10^{-8} \, \Omega^2RAl​RFe​=(0.3×10−4)×(12.5×10−4)=3.75×10−8Ω2
    • Equivalent resistance: Req=3.75×10−812.8×10−4=3.7512.8×10−4 ΩR_{eq} = {{3.75 \times 10^{-8}} \over {12.8 \times 10^{-4}}} = {{3.75} \over {12.8}} \times 10^{-4} \, \OmegaReq​=12.8×10−43.75×10−8​=12.83.75​×10−4Ω Req=37.5128×10−4=75256×10−4 ΩR_{eq} = {{37.5} \over {128}} \times 10^{-4} = {{75} \over {256}} \times 10^{-4} \, \OmegaReq​=12837.5​×10−4=25675​×10−4Ω

  7. Convert to Micro-ohms (μΩ\mu\OmegaμΩ) and Compare with Options:

    • Since 1 μΩ=10−6 Ω1 \, \mu\Omega = 10^{-6} \, \Omega1μΩ=10−6Ω, we have 1 Ω=106 μΩ1 \, \Omega = 10^6 \, \mu\Omega1Ω=106μΩ. Req=(75256×10−4)×106 μΩ=75×100256 μΩ=7500256 μΩR_{eq} = \left({{75} \over {256}} \times 10^{-4}\right) \times 10^6 \, \mu\Omega = {{75 \times 100} \over {256}} \, \mu\Omega = {{7500} \over {256}} \, \mu\OmegaReq​=(25675​×10−4)×106μΩ=25675×100​μΩ=2567500​μΩ
    • Simplify the fraction by dividing the numerator and denominator by 4: Req=7500÷4256÷4 μΩ=187564 μΩR_{eq} = {{7500 \div 4} \over {256 \div 4}} \, \mu\Omega = {{1875} \over {64}} \, \mu\OmegaReq​=256÷47500÷4​μΩ=641875​μΩ

This result matches option B.

Conclusion:

Evaluating the options: A: 247564μΩ{{2475} \over {64}}\mu \Omega642475​μΩ - Incorrect. B: 187564μΩ{{1875} \over {64}}\mu \Omega641875​μΩ - Correct. C: 187549μΩ{{1875} \over {49}}\mu \Omega491875​μΩ - Incorrect. D: 2475132μΩ{{2475} \over {132}}\mu \Omega1322475​μΩ - Incorrect.

Therefore, the only correct option is B.

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