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Standard XII

Physics

Semiconductors and Superconductors

Ends of two w...

Question

Ends of two wires $A$ and $B$ having resistivity $ρ_{A} = 3 \times 10^{- 5} Ω m$ and $ρ_{B} = 6 \times 10^{- 5} Ω m$ of same cross section area joined in series together to form a single wire. If the resistance of the joined wire does not change with temperature, then find the ratio of their lengths $\frac{l_{A}}{l_{B}}$ , given that temperature coefficient of resistivity of wire $A$ and $B$ is $α_{A} = 4 \times 10^{- 5} /^{\circ} C$ and $α_{B} = - 6 \times 10^{- 6} /^{\circ} C$ . Assume that mechanical dimensions do not change with temperature.

\frac{3}{7}

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\frac{10}{3}

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\frac{3}{10}

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\frac{1}{2}

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Solution

The correct option is C $\frac{3}{10}$
$R = R_{1} + R_{2}$
$R^{'} = R_{1}^{'} + R_{2}^{'}$
$R^{'} = R_{A} [1 + α_{A} Δ θ] + R_{B} [1 + α_{B} Δ θ]$
As net resistance does not change on changing temperature, so
$R_{A} α_{A} + R_{B} α_{B} = 0$
$ρ_{A} l_{A} α_{A} + ρ_{B} l_{B} α_{B} = 0$
$\frac{l_{A}}{l_{B}} = - \frac{α_{B}}{α_{A}} . \frac{ρ_{B}}{ρ_{A}}$
$\frac{l_{A}}{l_{B}} = \frac{3}{10}$

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The correct option is C 310R=R1+R2 R′=R′1+R′2 R′=RA[1+αAΔθ]+RB[1+αBΔθ] As net resistance does not change on changing temperature, so RAαA+RBαB=0 ρAlAαA+ρBlBαB=0 lAlB=−αBαA.ρBρA lAlB=310