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A copper wire and an aluminum wire have lengths in the ratio 1 : 2, diameters in the ratio 2 : 1 and forces applied in the ratio 4 : 3. Find the ratio of the increase in length of the two wires( $Y_{c u} = 1.1 \times 10^{11} N m^{- 2}$ $Y_{A l} = 0.70 \times 10^{11} N m^{- 2}$ )

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7 : 44

3 : 40

3 : 17

7 : 66

answer is A.

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Detailed Solution

$e = \frac{F l}{π r^{2} y}$

$\frac{e_{1}}{e_{2}} = \frac{F_{1}}{F_{2}} \times \frac{l_{1}}{l_{2}} \times {(\frac{r_{2}}{r_{1}})}^{2} = \frac{4}{3} \times (\frac{1}{2}) \times {(\frac{1}{2})}^{2} (\frac{0.70 \times 10^{11}}{1.1 \times 10^{11}}) = \frac{0.7}{6.6} = \frac{7}{66}$

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