Question

A wire of length 1m and radius of 1mm is subjected to a load. The extension is x. The wire is melted and drawn into a wire of square cross section of side 2mm. Its extension under the same load will be
$\begin{aligned} & a)\dfrac{{{\pi }^{2}}x}{8} \\\ & b)\dfrac{{{\pi }^{2}}x}{16} \\\ & c)\dfrac{{{\pi }^{2}}x}{2} \\\ & d)\dfrac{x}{2\pi } \\\ \end{aligned}$

Answer 1

It is given in the question that the wire of the same material is melted into a wire of square cross section. Hence the Young’s modulus for the wire will be the same. The Young’s modulus (Y)for a wire subjected to a load is given by, $\text{Y =}\dfrac{Load/A}{\Delta L/L}$, where load is basically the force applied on the wire with the area of cross section A, $\Delta L$, is the extension in the wire and L is its original length before extension. Hence we can equate the above equation for both the wires and determine the change in length of the wire with a square cross section.

Complete step by step answer:
To begin with the question let us first find the length of the wire when melted into a wire of square cross section. When a wire is melted into any shape the volume remains constant. Hence we can write for the above two wire,
${{V}_{CIRCULAR}}={{V}_{SQUARE}}$
where ${{V}_{CIRCULAR}}$ is the volume of the wire with circular cross section and ${{V}_{SQUARE}}$ is the volume of the wire with square cross section.
The above equation can also be written as,
$\begin{aligned} & {{V}_{CIRCULAR}}={{V}_{SQUARE}} \\\ & \pi {{r}^{2}}l=AL, \\\ \end{aligned}$
Where l is the length of the wire with circular and L is the length of the wire with square cross section. Substituting the respective measurements in the above equation we get,
$\begin{aligned} & \Rightarrow \pi {{(1\times {{10}^{-3}})}^{2}}\times 1m={{(2\times {{10}^{-3}})}^{2}}L \\\ & \Rightarrow \pi (1\times {{10}^{-6}})=(4\times {{10}^{-6}})L \\\ & \Rightarrow L=\dfrac{\pi }{4}m \\\ \end{aligned}$
Hence the length of the wire with the square cross section is $\dfrac{\pi }{4}$. Hence we can write the Young’s modulus for wire with circular cross section when the extension in the wire is x as,
$\begin{aligned} & \text{Y =}\dfrac{Load/\pi {{r}^{2}}}{x/l} \\\ & \text{Y=}\dfrac{Load/\pi {{\left( {{10}^{-3}} \right)}^{2}}}{x/1m} \\\ & \Rightarrow \text{Y =}\dfrac{Load}{\pi \times {{10}^{-6}}\times x}...(1) \\\ \end{aligned}$
Let us denote the extension in the wire with a square cross section as y. Therefore the Young’s modulus for that can be written as,
$\begin{aligned} & \text{Y=}\dfrac{Load/{{A}^{2}}}{y/L} \\\ & \text{Y=}\dfrac{Load/{{\left( 2\times {{10}^{-3}} \right)}^{2}}}{y/\dfrac{\pi }{4}m} \\\ & \Rightarrow \text{Y =}\dfrac{Load\times \dfrac{\pi }{4}}{4\times {{10}^{-6}}\times y}...(2) \\\ \end{aligned}$
Since the Young’s modulus for both the wires is the same, therefore lets equate equation 1 to 2.
$\dfrac{Load}{\pi \times {{10}^{-6}}\times x}=\dfrac{Load\times \dfrac{\pi }{4}}{4\times {{10}^{-6}}\times y}$
Both the wires are subjected to same load, hence
$\begin{aligned} & \dfrac{1}{\pi \times x}=\dfrac{\pi }{16\times y} \\\ & \therefore y=\dfrac{{{\pi }^{2}}x}{16} \\\ \end{aligned}$
Hence the correct answer is option b.

Note:
We have assumed that the volume of the wire remains the same when melted. Hence it is also one of the factors because of which the Young’s modulus for the material can be equated. The term load upon area is nothing but the stress on the wire i.e. pressure.