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Mechanical Properties of Solids

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Question : 10 of 21
Marks: +1, -0
A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 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.
[Young’s modulus of elasticity for copper and steel are 110×109 N m−2110 \times 10^{9} \,\mathrm{N}\,\mathrm{m}^{-2} and 190×109 N m−2190 \times 10^{9} \,\mathrm{N}\,\mathrm{m}^{-2} respectively.]
Solution:  
As each wire has same tension F, so each wire has same extension due to mass of rigid bar. As each wire is of same length, hence each wire has same strain. If D is the diameter of wire, then
Y=4FπD2strainY = \frac{\frac{4F}{\pi D^{2}}}{\text{strain}} or D2∝1YD^{2} \propto \frac{1}{Y}
DCuDiron=YironYCu\frac{D_{\mathrm{Cu}}}{D_{\mathrm{iron}}} = \sqrt{\frac{Y_{\mathrm{iron}}}{Y_{\mathrm{Cu}}}}
=190×109110×109= \sqrt{\frac{190 \times 10^{9}}{110 \times 10^{9}}}
=1911=1.31= \sqrt{\frac{19}{11}} = 1.31
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