Three copper wires of length and cross-sectional area are l-A- 2l-A2 and l2- 2A and their resistance are R1 - R2 and R3 respectively- The increasing order of resistances is-

Given, l_1=l ; l_2=2l ; l_3=l2 A_1=A ; A_2=A2 ; A_3=2A Resistance of a wire is, R= ρ lA ⇒ R_1 = ρ lA=R ⇒ R_2 = ρ 2lA2=4(ρ lA)=4R ⇒ R_3 = ρ l2(2 ...

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Standard XII

Physics

Series and Parallel Combination of Resistors

Three copper ...

Question

Three copper wires of length and cross-sectional area are $(l, A)$ , $(2 l, \frac{A}{2})$ and $(\frac{l}{2}, 2 A)$ and their resistance are $R_{1}, R_{2}$ and $R_{3}$ respectively. The increasing order of resistances is-

R_{1} > R_{2} > R_{3}

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R_{3} > R_{2} > R_{1}

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R_{2} > R_{3} > R_{1}

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R_{3} < R_{1} < R_{2}

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Solution

The correct option is D $R_{3} < R_{1} < R_{2}$
Given, $l_{1} = l; l_{2} = 2 l; l_{3} = \frac{l}{2}$

$A_{1} = A; A_{2} = \frac{A}{2}; A_{3} = 2 A$

Resistance of a wire is,

$R = \frac{ρ l}{A}$

$\Rightarrow R_{1} = \frac{ρ l}{A} = R$

$\Rightarrow R_{2} = \frac{ρ 2 l}{\frac{A}{2}} = 4 (\frac{ρ l}{A}) = 4 R$

$\Rightarrow R_{3} = \frac{ρ l}{2 (2 A)} = \frac{1}{4} (\frac{ρ l}{A}) = \frac{R}{4}$

$∴ R_{3} < R_{1} < R_{2}$

 Hence, $(D)$ is the correct answer.

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