Question

A rigid bar of mass $M$ is supported symmetrically by three wires each of length $L$. Those at each end are of copper and the middle one is of iron. The ratio of their diameters, if each is to have the same tension, is equal to

• A. $\frac{\Upsilon_{copper}}{\Upsilon_{iron}}$
• B. $\sqrt{\frac{\Upsilon_{iron}}{\Upsilon_{copper}}}$
• C. $\frac{\Upsilon^2_{iron}}{\Upsilon^2_{copper}}$
• D. $\frac{\Upsilon_{iron}}{\Upsilon_{copper}}$

Answer 1

Answer: (B)

$\sqrt{\frac{\Upsilon_{iron}}{\Upsilon_{copper}}}$

Answer 2

As $\Upsilon=\frac{FL}{A\Delta L}=\frac{FL}{\left(\pi D^{2}/4\right)\Delta L}$ $=\frac{4FL}{\pi D^{2}\,\Delta L}$ For same tension $\left(F\right)$, $\Upsilon \propto \frac{1}{D^{2}}$ or $D\propto\frac{1}{\sqrt{\Upsilon}}$ or $\frac{D_{copper}}{D_{iron}}=\sqrt{\frac{\Upsilon_{iron}}{\Upsilon_{copper}}}$