A rigid bar of mass 15; kg is supported symme | vedclass

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(A) The tension force $F$ acting on each wire is the same. Since the wires are of the same length $L$,the extension $\Delta L$ in each case is the sam

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Ratio of diameter of copper to iron wire is $1.31:1$

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(A) The tension force $F$ acting on each wire is the same. Since the wires are of the same length $L$,the extension $\Delta L$ in each case is the same. Thus,the strain $\epsilon = \frac{\Delta L}{L}$ is the same for all wires.From the definition of Young's modulus $Y = \frac{	ext{Stress}}{	ext{Strain}} = \frac{F/A}{\epsilon}$,we have $A = \frac{F}{Y \epsilon}$.Since $F$ and $\epsilon$ are constant,$A \propto \frac{1}{Y}$.Given the area of a circular wire $A = \frac{\pi d^2}{4}$,we have $d^2 \propto \frac{1}{Y}$,or $d \propto \frac{1}{\sqrt{Y}}$.Therefore,the ratio of the diameter of the copper wire $(d_c)$ to the iron wire $(d_i)$ is $\frac{d_c}{d_i} = \sqrt{\frac{Y_i}{Y_c}}$.Given $Y_i = 190 	imes 10^9 \; Pa$ and $Y_c = 110 	imes 10^9 \; Pa$,we get:$\frac{d_c}{d_i} = \sqrt{\frac{190 	imes 10^9}{110 	imes 10^9}} = \sqrt{\frac{19}{11}} \approx 1.31$.Thus,the ratio of their diameters is $1.31:1$.

$A$ rigid bar of mass $15\; kg$ is supported symmetrically by three wires each $2.0\; m$ long. Those 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.

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