copper wire and an aluminium 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. $(Y_{Cu} = 1.1 \times 10^{11} {Nm}^{2}, Y_{Al} = 0.70 \times 10^{11} {Nm}^{- 2})$

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110 : 189

180 : 110

189 : 110

80 : 11

answer is C.

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Detailed Solution

$\frac{∆ l_{1}}{∆ l_{2}} = \frac{F_{1}}{F_{2}} \times \frac{l_{1}}{l_{2}} \times \frac{{r_{2}}^{2}}{{r_{1}}^{2}} \times \frac{Y_{2}}{Y_{1}}$

$\frac{∆ l_{1}}{∆ l_{2}} = (\frac{4}{5}) (\frac{3}{2}) (\frac{0.7}{1.1}) (\frac{9}{4}) = \frac{189}{110}$

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