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. $\frac{Y_{copper}}{Y_{iron}}$
• B. $\sqrt{\frac{Y_{iron}}{Y_{copper}}}$
• C. $\frac{Y_{iron}^2}{Y_{copper}^2}$
• D. $\frac{Y_{iron}}{Y_{copper}}$

Answer 1

Answer: (B)

The situation is as shown in the figure.
Let $T$ be tension in each wire.
As the bar is supported symmetrically by the three wires, therefore extension in each wire is same.
As $Y=\frac{F/A}{\Delta L/L}$
If D is the diameter of the wire,

then $Y=\frac{F/\pi {\left(D/2\right)}^2}{\Delta L/L}$ $=\frac{4FL}{\pi D^2\Delta L}$
As per the conditions of the problem, $F\left(tension\right)$, length $L$, and extension $\Delta L$ is same for each wire.
$\therefore$ $Y\propto \frac{1}{D^2}$ or $D\propto \sqrt{\frac{1}{Y}}$ $\therefore$ $\frac{D_{copper}}{D_{iron}}$ $=\sqrt{\frac{Y_{iron}}{Y_{copper}}}$