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Current Electricity question

2013 · Q124
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Current Electricity question

2013 · Q124

NEETPhysicsCurrent ElectricityMCQ+4 / −1
Two rods are joined end to end, as shown. Both have a cross-sectional area of 0.01 cm2. Each is 1 meter long. One rod is of copper with a resistivity of 1.7 ×\times× 10−-−6 ohm-centimeter, the other is of iron with a resistivity of 10−-−5 ohm-centimeter.
NEET 2013 (Karnataka) Physics - Current Electricity Question 92 English
How much voltage is required to produce a current of 1 ampere in the rods?
  1. A
    0.00145 V
  2. B
    0.0145 V
  3. C
    1.7 ×\times× 10−-−6 V
  4. D
    0.117 V
View written solutionFree

Correct answer: D

Copper rod and iron rod are joined in series.

∴R=RCu+RFe=(ρ1+ρ2)ℓA \therefore R = {R_{Cu}} + {R_{Fe}} = \left( {{\rho _1} + {\rho _2}} \right){\ell \over A}∴R=RCu​+RFe​=(ρ1​+ρ2​)Aℓ​
(∵R=ρℓA)\left(\because {R = \rho {\ell \over A}} \right)(∵R=ρAℓ​)

From ohm’s law
V = RI = (1.7 × 10–6 × 10–2 + 10–5 × 10–2)
÷\div÷ 0.01 × 10–4 volt

= 0.117 volt   (∵\because∵ I = 1A)

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