A rigid bar of mass 15 , kg is supported symm | vedclass

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(C) Given that the tension $T$ in each wire is the same.From the free body diagram of the bar,the total upward force is $3T$ and the downward force is

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$\sqrt{\frac{19}{11}}$

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(C) Given that the tension $T$ in each wire is the same.From the free body diagram of the bar,the total upward force is $3T$ and the downward force is $mg = 15 	imes 10 = 150 \, N$.Thus,$3T = 150 \, N \Rightarrow T = 50 \, N$.Since the bar is supported symmetrically,the extension $\Delta L$ in each wire must be the same.We know that the extension $\Delta L = \frac{FL}{AY}$,where $F$ is the tension,$L$ is the length,$A$ is the cross-sectional area,and $Y$ is the Young's modulus.Since $F$,$L$,and $\Delta L$ are the same for all wires,we have $A_C Y_C = A_S Y_S$,where $C$ denotes copper and $S$ denotes steel.Therefore,$\frac{A_C}{A_S} = \frac{Y_S}{Y_C}$.Substituting $A = \frac{\pi d^2}{4}$,we get $\frac{d_C^2}{d_S^2} = \frac{Y_S}{Y_C}$.$\frac{d_C}{d_S} = \sqrt{\frac{Y_S}{Y_C}} = \sqrt{\frac{190 	imes 10^9}{110 	imes 10^9}} = \sqrt{\frac{19}{11}}$.

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