An Aluminium and Copper wire of same cross sectional area but having lengths in the ratio $2 : 3$ are joined end to end. This composite wire is hung from a rigid support and a load is suspended from the free end. If the increase in length of the composite wire is $2.1 m m$ , the increase in lengths of Aluminium and Copper wires are : [ $Y_{A l} = 20 \times 10^{11} N / m^{2}$ and $Y_{C u} = 12 \times 10^{11} N / m^{2}$ ]

0.7 m m; 1.4 m m

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0.9 m m; 1.2 m m

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1.0 m m; 1.1 m m

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0.6 m m; 1.5 m m

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Solution

The correct option is C $0.6 m m; 1.5 m m$
Since the tension and cross section area of both the wires are same, stress in both the wires is same.

Since, $\frac{T}{A} = Y \times \frac{Δ l}{l}$

Hence, $Δ l = k \frac{l}{Y}$

$\frac{Δ l_{1}}{Δ l_{2}} = \frac{l_{1}}{l_{2}} \times \frac{Y_{2}}{Y_{1}} = \frac{2}{3} \times \frac{12}{20} = \frac{2}{5}$

Hence, $2 x + 5 x = 2.1 m m$ (Given)

$x = 0.3 m m$

Hence the lengths are $2 x = 0.6 m m$ and $5 x = 1.5 m m$

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The correct option is C 0.6 mm;1.5 mmSince the tension and cross section area of both the wires are same, stress in both the wires is same.Since, TA=Y×ΔllHence, Δl=klYΔl1Δl2=l1l2×Y2Y1=23×1220=25Hence, 2x+5x=2.1 mm (Given)x=0.3 mmHence the lengths are 2x=0.6 mm and 5x=1.5 mm