Question

A composite wire of uniform diameter of $3.0mm$ consisting of a copper wire of length $2.2m$ and a steel wire of length $1.6m$ stretches under a load by $0.7mm$ . Calculate the load, given that the Young’s modulus for copper is $1.1 \times {10^{11}}Pa$ and for steel is $2.0 \times {10^{11}}Pa$ .

Answer 1

Hint: Start with finding the radius of the wire then find the change in length of both the copper and steel wire when the load is applied. After that use the formula of the force in terms of young’s modulus and length of the wire and put all the values in the formula and get the required answer.




Complete answer:
According to the question,
The diameter of the wire is $3.0mm$.
So, the radius of wire will be, $r = \dfrac{3}{2}mm = 1.5 \times {10^{ - 3}}m$
Since the area of cross-section A of each wire is exactly the same, the stress is the same for both wires.
$Stress=\dfrac {F}{A}=y\times{strain}=y\dfrac {\Delta~l}{l}$
Now the force per unit area will be equal for copper and steel:
We know, $Y=\frac{F/A}{\Delta~l/l}$
Therefore, $\dfrac{F}{A} = Y \times \dfrac{{\Delta l}}{l}$
Equating for both the copper and steel, we get;
${Y_{Cu}} \times \dfrac{{\Delta {l_{Cu}}}}{{{l_{Cu}}}} = {Y_{St}} \times \dfrac{{\Delta {l_{St}}}}{{{l_{St}}}}$
Rearranging the above equation and outing values from the question, we get;
$\dfrac{{\Delta {l_{Cu}}}}{{\Delta {l_{St}}}} = \dfrac{{{Y_{St}} \times {l_{Cu}}}}{{{Y_{Cu}} \times {l_{St}}}} = \dfrac{{2 \times {{10}^{11}} \times 2.2}}{{1.1 \times {{10}^{11}} \times 1.6}}$
By solving, we get;
$\Delta {l_{Cu}} = 2.5\Delta {l_{St}}$
As $\Delta {l_{Cu}} + \Delta {l_{St}} = 0.7 \times {10^{ - 3}}$
Now, $2.5\Delta {l_{st}} + \Delta {l_{St}} = 0.7 \times {10^{ - 3}}$
$3.5\Delta {l_{St}} = 0.7 \times {10^{ - 3}}$
On solving $\Delta {l_{St}} = 2.0 \times {10^{ - 4}}$
Therefore, $\Delta {l_{Cu}} = 2.5\Delta {l_{St}} = 2.5 \times 2 \times {10^{ - 4}} = 5 \times {10^{ - 4}}$
We know that;
$\dfrac{F}{A} = Y\dfrac{{\Delta l}}{l}$
Therefore, $F = \dfrac{{A \times Y \times \Delta l}}{l}$
Taking value for copper wire, $F = \dfrac{{\pi {r^2} \times {Y_{Cu}} \times \Delta {l_{Cu}}}}{{{l_{Cu}}}}$
Putting all the values, we get;
$F = \dfrac{{3.14{{\left( {1.5 \times {{10}^{ - 3}}} \right)}^2} \times 1.1 \times {{10}^{11}} \times 5 \times {{10}^{ - 4}}}}{{2.2}} = 176.8N$
The load will be 176.8N.




Note: Be careful about the unit of the quantity given, change all the quantities in the same unit while putting in the formula and then solve it otherwise you will get the wrong answer. Also, use the correct formula for the load applied here, it was in terms of area, Young’s modulus, and the change in length.