A copper wire and an aluminum wire have lengths in the ratio 3 : 2, diameters in the ratio 2 : 3 and forces applied in the ratio 4 : 5. 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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$110 : 189$

$180 : 110$

$189 : 110$

$80 : 11$

answer is C.

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Detailed Solution

$Y = \frac{F ℓ}{π r^{2} e} \Rightarrow e = \frac{F ℓ}{π r^{2} Y} \Rightarrow \frac{e_{1}}{e_{2}} = \frac{F_{1}}{F_{2}} \times \frac{l_{1}}{l_{2}} \times {(\frac{r_{2}}{r_{1}})}^{2} \times \frac{Y_{2}}{Y_{1}} \Rightarrow \frac{e_{1}}{e_{2}} = \frac{4}{5} \times \frac{3}{2} \times {(\frac{3}{2})}^{2} \times \frac{0.7 \times 10^{11}}{1.1 \times 10^{11}} = \frac{189}{110}$