A copper wire of radius r and length 1 has a resistance of R. A second

A copper wire of radius r and length 1 has a resistance of R. A second copper wire with radius 2r and length 1 is taken and the two wires are joined in a parallel combination. The resultant resistance of the parallel combination of the two wires will be

[amp_mcq option1=”5R” option2=”5/4 R” option3=”4/5 R” option4=”R/5″ correct=”option4″]

This question was previously asked in

UPSC CDS-1 – 2017

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The resistance of a wire is given by the formula R = ρ * (l/A), where ρ is the resistivity of the material, l is the length, and A is the cross-sectional area. The area of a circular wire is A = πr², where r is the radius.

– For the first wire: R₁ = R = ρ * (l / πr²).
– For the second wire: The radius is 2r, and the length is l. So, R₂ = ρ * (l / π(2r)²) = ρ * (l / 4πr²) = (1/4) * [ρ * (l / πr²)] = R/4.
– When two resistors R₁ and R₂ are connected in parallel, the resultant resistance R_eq is given by 1/R_eq = 1/R₁ + 1/R₂ or R_eq = (R₁ * R₂) / (R₁ + R₂).
– Plugging in the values, R_eq = (R * (R/4)) / (R + R/4) = (R²/4) / (5R/4).
– R_eq = (R²/4) * (4/5R) = R/5.

This problem illustrates the relationship between resistance, material properties, and geometry, as well as the calculation of equivalent resistance in a parallel circuit. The resistivity (ρ) is the same for both wires as they are both copper.

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