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Three copper wires of lengths and cross-sectional areas are (l, A), (2l, A/2), and (l/2, 2A). Resistance is minimum in
1. wire of cross-sectional area 2A
2. wire of cross-sectional area \(\frac{A}{2}\)
3. wire of cross-sectional area A
4. the same in all three cases

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Correct Answer - Option 1 : wire of cross-sectional area 2A

Concept:

The resistance offered to the flow of current is known as the resistance.

SI unit of resistance is the ohm (Ω).

Mathematically, the resistance can be written as:

\(R \propto \frac{l}{A}\)

\(R = \frac{{\rho l}}{A}\)

R = resistance

l = length

A = area of cross-section

ρ = resistivity

Calculation:

Three wires of different lengths and cross-sectional areas are given as:

(l, A), (2l, A/2), and (l/2, 2A).

For Ist wire:

R1 ∝ l / A = R

For IInd wire:

\({R_2} \propto \frac{{2l}}{{A/2}} = 4R\)

For IIIrd wire:

\({R_3} \propto \frac{{l/2}}{{2A}} = \frac{R}{4}\)

Therefore resistance of the wire will be minimum for IIIrd wire.

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