A copper wire and an aluminium wire have lengths in the ratio 5 : 2, diameters in the ratio 4 : 3 and forces applied in the ratio 4 : 5. Find the ratio of increase in length of the copper wire to that of aluminium wire.(Let YCu=1.1×1011 Nm-2, YAl=0.7×1011 Nm-2 )

# A copper wire and an aluminium wire have lengths in the ratio 5 : 2, diameters in the ratio 4 : 3 and forces applied in the ratio 4 : 5. Find the ratio of increase in length of the copper wire to that of aluminium wire.

1. A

$\frac{176}{63}$

2. B

$\frac{63}{88}$

3. C

$\frac{189}{110}$

4. D

$\frac{33}{89}$

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### Solution:

Given the ratio of copper to aluminum wire length is ${l}_{1}:{l}_{2}=5:2$

and the wire's diameter ratio is ${\mathrm{d}}_{1}:{\mathrm{d}}_{2}=4:3$

The increase in length of a wire under tension can be calculated using Hooke's Law:

copper wire,

$∆{l}_{1}=\frac{{\mathrm{F}}_{1}{l}_{1}}{{\mathrm{A}}_{1}{\mathrm{Y}}_{1}}\mathrm{________}\left(1\right)$

aluminum wire,

$∆{l}_{2}=\frac{{\mathrm{F}}_{2}{l}_{2}}{{\mathrm{A}}_{2}{\mathrm{Y}}_{2}}\mathrm{________}\left(2\right)$

from equations 1 and 2,

$\frac{∆{l}_{1}}{∆{l}_{2}}=\frac{{l}_{1}}{{l}_{2}}.\frac{{\mathrm{A}}_{2}}{{\mathrm{A}}_{1}}.\frac{{\mathrm{Y}}_{2}}{{\mathrm{Y}}_{1}}.\frac{{\mathrm{F}}_{1}}{{\mathrm{F}}_{2}}\left[\frac{{\mathrm{Y}}_{2}}{{\mathrm{Y}}_{1}}=\frac{{\mathrm{Y}}_{\mathrm{Al}}}{{\mathrm{Y}}_{\mathrm{Cu}}}=\frac{0.7}{1.1}\right]$

$\frac{∆{l}_{1}}{∆{l}_{2}}=\frac{5}{2}.\frac{{{\mathrm{\pi r}}_{2}}^{2}}{{{\mathrm{\pi r}}_{1}}^{2}}.\frac{0.7×{10}^{11}}{1.1×{10}^{11}}.\frac{4}{5}$

$\frac{∆{l}_{1}}{∆{l}_{2}}=\frac{5}{2}×\frac{{\left(\frac{3}{2}\right)}^{2}}{{\left(2\right)}^{2}}×\frac{7}{11}×\frac{4}{5}$

$\frac{∆{l}_{1}}{∆{l}_{2}}=\frac{63}{88}$

Hence the correct answer is $\frac{63}{88}.$

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