Question

A 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 (${Y_{Cu}} = 1.1 \times {10^{11}}\;{\rm{N}}{{\rm{m}}^{ - 2}}$, ${Y_{Al}} = 0.70 \times {10^{11}}\;{\rm{N}}{{\rm{m}}^{ - 2}}$).
(1) $110:189$
(2) $180:110$
(3) $189:110$
(4) $80:11$

Answer 1

Here, we will use the formula for Young’s modulus of a material. Young’s modulus is the ratio of stress to strain and the formula is derived using this definition.

Complete step by step answer:
The ratio of the length of copper wire to aluminium wire, ${l_{Cu}}:{l_{Al}} = 3:2$
The ratio of the diameter of copper wire to aluminium wire, ${d_{Cu}}:{d_{Al}} = 2:3$
The ratio of the forces applied on the copper wire to the aluminium wire, ${F_{Cu}}:{F_{Al}} = 4:5$
The Young’s modulus of the copper wire, ${Y_{Cu}} = 1.1 \times {10^{11}}\;{\rm{N}}{{\rm{m}}^{ - 2}}$
The Young’s modulus of the aluminium wire, ${Y_{Al}} = 0.70 \times {10^{11}}\;{\rm{N}}{{\rm{m}}^{ - 2}}$
The Young’s modulus of a wire can be written as
$Y = \dfrac{\sigma }{\varepsilon }$
Here $\sigma$ is the stress on the wire and $\varepsilon$ is the strain on the wire.
Now, the relation for stress on the wire can be written as
$\sigma = \dfrac{F}{A}$
Here $F$ is the force applied on the wire and $A$ is the area of the cross section of the wire.
We can write the area of cross section of the wire as
$A = \pi {r^2}$
Here $r$ is the radius of the wire.
Hence, the stress equation becomes
$\sigma = \dfrac{F}{{\pi {r^2}}}$
Now, the strain on the wire can be written as
$\varepsilon = \dfrac{{\Delta l}}{l}$
Here $\Delta l$ is the increase in length of the wire and $l$ is the original length of the wire.
We can substitute the relations for stress and strain in the Young’s modulus equation. Then, the equation for Young’s modulus becomes
$Y = \dfrac{{\dfrac{F}{{\pi {r^2}}}}}{{\dfrac{{\Delta l}}{l}}}\\\ = \dfrac{{Fl}}{{\pi {r^2}\Delta l}}$
Now, from this equation we can obtain the relation for the change in length as
$\Delta l = \dfrac{{Fl}}{{Y\pi {r^2}}}$
Using this equation, we can write the ratio of the increase in length of the copper wire to the aluminium wire as
$\dfrac{{\Delta {l_{Cu}}}}{{\Delta {l_{Al}}}} = \dfrac{{\dfrac{{{F_{Cu}}{l_{Cu}}}}{{{Y_{Cu}}\pi r_{Cu}^2}}}}{{\dfrac{{{F_{Al}}{l_{Al}}}}{{{Y_{Al}}\pi r_{Al}^2}}}}\\\ = \dfrac{{{F_{Cu}}{l_{Cu}}{Y_{Al}}r_{Al}^2}}{{{F_{Al}}{l_{Al}}{Y_{Cu}}r_{Cu}^2}}\\\ \implies \dfrac{{\Delta {l_{Cu}}}}{{\Delta {l_{Al}}}} = \dfrac{{{F_{Cu}}}}{{{F_{Al}}}} \times \dfrac{{{I_{Cu}}}}{{{l_{Al}}}} \times \dfrac{{{Y_{Al}}}}{{{Y_{Cu}}}} \times {\left( {\dfrac{{{r_{Al}}}}{{{r_{Cu}}}}} \right)^2}$
It is given that $\dfrac{{{F_{Cu}}}}{{{F_{Al}}}} = \dfrac{4}{5}$, $\dfrac{{{I_{Cu}}}}{{{l_{Al}}}} = \dfrac{3}{2}$ and $\dfrac{{{d_{Cu}}}}{{{d_{Al}}}} = \dfrac{2}{3}$.
Since the ratio of the diameters and the ratio of radii will be the same, we can write
$\dfrac{{{r_{Cu}}}}{{{r_{Al}}}} = \dfrac{2}{3}$
Now we substitute the values for ${F_{Cu}}$, ${F_{Al}}$, $\dfrac{{{F_{Cu}}}}{{{F_{Al}}}}$, $\dfrac{{{I_{Cu}}}}{{{l_{Al}}}}$and $\dfrac{{{r_{Cu}}}}{{{r_{Al}}}}$ in the equation for $\dfrac{{\Delta {l_{Cu}}}}{{\Delta {l_{Al}}}}$ to get

$$\dfrac{{\Delta {l_{Cu}}}}{{\Delta {l_{Al}}}} = \dfrac{4}{5} \times \dfrac{3}{2} \times \dfrac{{0.70 \times {{10}^{11}}\;}}{{1.1 \times {{10}^{11}}\;}} \times {\left( {\dfrac{3}{2}} \right)^2}\\\ = \dfrac{4}{5} \times \dfrac{3}{2} \times \dfrac{7}{{11}} \times \dfrac{9}{4}\\\ = \dfrac{{189}}{{110}}$$

Therefore, the option (3) is the correct answer.

Note:
Be careful to insert the correct ratios into the equation. In the final equation, some ratios are expressed as copper to aluminium and others as aluminium to copper. We should take care to inverse the given ratios if required.