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Home » A rigid bar of mass is supported symmetrically by three wires each long. Those at each end are of copper and the middle one is of iron. Determine the ratios of their diameters if each is to have the same tension.
A rigid bar of mass 15 \mathrm{~kg} is supported symmetrically by three wires each 2.0 \mathrm{~m} long. Those at each end are of copper and the middle one is of iron. Determine the ratios of their diameters if each is to have the same tension.
CBSE | Class 11 | Mechanical Properties of Solids | NCERT | Physics
A rigid bar of mass 15 \mathrm{~kg} is supported symmetrically by three wires each 2.0 \mathrm{~m} long. Those at each end are of copper and the middle one is of iron. Determine the ratios of their diameters if each is to have the same tension.

The extension of each wire will be the same because the stress on the wires is the same. Now that the wires are the same length, the tension on them will be the same.
Now, we know:

Y=Stress / Strain

=(\mathrm{F} / \mathrm{A}) / \mathrm{Strain}
=\left(4 \mathrm{~F} / \mathrm{m} \mathrm{d}^{2}\right) / \mathrm{Strain} \quad \ldots \ldots \ldots(1)

From equation (1), We can conclude that \mathrm{Y} \propto\left(1 / \mathrm{d}^{2}\right)

We know that Young’s modulus for iron is Y_{1}=190 \times 10^{9} \mathrm{~Pa}

Let d_{1} be the diameter of the iron wire

Also, Young’s modulus for copper is known as Y_{2}=120 \times 10^{9} \mathrm{~Pa}

let \mathrm{d}_{2} be the diameter of the copper wire

Thus, the ratio of their diameters can be given as :

\frac{d_{1}}{d_{2}}=\sqrt{\frac{Y_{1}}{Y_{2}}}

i

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