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A rigid bar of mass $15 \mathrm{~kg}$ is supported symmetrically by three wires each $2 \mathrm{~m}$ long. The wires at each end are of copper and middle one is of iron. Determine the ratio of their diameters if each has the same tension. Young's modulus of elasticity for copper and steel are $110 \times 10^9$ $\mathrm{N} / \mathrm{m}^2$ and $190 \times 10^9 \mathrm{~N} / \mathrm{m}^2$ respectively.
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Each wire has same tension so each wire will have same extension. Also as they have same length, each wire will have same strain
$$
\begin{aligned}
&Y=\frac{F l}{A \Delta l}=\frac{F l}{\pi(D / 2)^2 \Delta l}=\frac{4 F l}{\pi D^2 \Delta l} \\
&\therefore D^2 \propto \frac{1}{Y} \\
&\therefore \frac{D_{\mathrm{Cu}}}{D_{\text {Steel }}}=\sqrt{\frac{Y_{\text {Steel }}}{Y_{\mathrm{Cu}}}}=\sqrt{\frac{190 \times 10^9}{110 \times 10^9}}=\sqrt{\frac{19}{11}}=1.31
\end{aligned}
$$
$$
\begin{aligned}
&Y=\frac{F l}{A \Delta l}=\frac{F l}{\pi(D / 2)^2 \Delta l}=\frac{4 F l}{\pi D^2 \Delta l} \\
&\therefore D^2 \propto \frac{1}{Y} \\
&\therefore \frac{D_{\mathrm{Cu}}}{D_{\text {Steel }}}=\sqrt{\frac{Y_{\text {Steel }}}{Y_{\mathrm{Cu}}}}=\sqrt{\frac{190 \times 10^9}{110 \times 10^9}}=\sqrt{\frac{19}{11}}=1.31
\end{aligned}
$$
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