Question

A rigid bar of mass 15 kg is supported symmetrically by three wires each 2 m long. These at each end are of copper and the middle one is of iron. Determine the ratio of their diameters if each is to have the same tension.

Answer 1

The question is based on the concept of Young’s modulus of the material. This problem can be easily solved by keeping all the other parameters, except the area as constant, as the ratio of the diameters of the wires is being asked.
Formula used:
$Y=\dfrac{Stress}{Strain}$

Complete answer:
Consider Young’s modulus formula
As Young’s modulus is a ratio of stress by strain, so, we have,
$Y=\dfrac{Stress}{Strain}$
Where stress is the ratio of force by area and the strain is the elongation/change in the length.

So, we have,
$Y=\dfrac{{}^{F}/{}_{A}}{{}^{\Delta l}/{}_{l}}$
From the given question statement, it’s clear that the elongation remains constant, that is, the strain remains the same for all the three wires, as all the wires are supposed to have the same tension.

As the tension remains the same, thus, the force, which is related to the tension also remains the same.
Now let us consider the parameter A, that is the area.
The formula for the area can be written as follows.
$A=\pi {{r}^{2}}$
This formula can be represented in terms of the diameter as follows.

& A=\pi {{\left( \dfrac{d}{2} \right)}^{2}} \\\ & \Rightarrow A=\dfrac{\pi {{d}^{2}}}{4} \\\ \end{aligned}$$ As of now, consider Young’s modulus of the copper and the iron wires. So, we have, The Young’s modulus of the copper is given as follows. $${{Y}_{c}}=\dfrac{{}^{F}/{}_{{{A}_{c}}}}{{}^{\Delta l}/{}_{l}}$$ …… (1) Similarly, Young’s modulus of the iron is given as follows. $${{Y}_{I}}=\dfrac{{}^{F}/{}_{{{A}_{I}}}}{{}^{\Delta l}/{}_{l}}$$…… (2) Divide the equations (1) and (2), so, we get, $$\dfrac{{{Y}_{C}}}{{{Y}_{I}}}=\dfrac{{{A}_{I}}}{{{A}_{C}}}$$ Now represent the area in terms of the diameter of wire. So, we get, $$\begin{aligned} & \dfrac{{{Y}_{C}}}{{{Y}_{I}}}=\dfrac{{}^{\pi {{d}^{2}}_{I}}/{}_{4}}{{}^{\pi {{d}^{2}}_{C}}/{}_{4}} \\\ & \Rightarrow \dfrac{{{Y}_{C}}}{{{Y}_{I}}}=\dfrac{{{d}^{2}}_{I}}{{{d}^{2}}_{C}} \\\ \end{aligned}$$ Now rearrange the terms, in order to express the formula in terms of the diameter of the wire. Thus, we get, $$\dfrac{{{d}_{C}}}{{{d}_{I}}}=\sqrt{\dfrac{{{Y}_{I}}}{{{Y}_{C}}}}$$ Now substitute Young’s modulus values of the copper and iron wires in the above equation to find the ratio of the diameters. $$\begin{aligned} & \dfrac{{{d}_{C}}}{{{d}_{I}}}=\sqrt{\dfrac{190\times {{10}^{9}}}{120\times {{10}^{9}}}} \\\ & \Rightarrow \dfrac{{{d}_{C}}}{{{d}_{I}}}=\dfrac{1}{1.25} \\\ \end{aligned}$$ Therefore, the ratio of the diameters of the copper and iron wire is 1 : 1.25. **Note:** The things to be on your finger-tips for further information on solving these types of problems are: To confuse the students, the value of the mass of a rigid body is given, but it is of no use while calculating the ratio of the diameters of the wires.