A rigid bar of mass $15\,kg$ is supported symmetrically by three wire each of $2 \,m$ long. These at each end are of copper and middle one is of steel. Young's modulus of elasticity for copper and steel are $110 \times 10^9 \,N / m ^2$ and $190 \times 10^9 \,N / m ^2$ respectively. If each wire is to have same tension, ratio of their diameters will be $ ............$
Medium
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Tension is same $($given$)$
From free body diagram
$3 T=150 \,N$
$T=50 \,N$
Since the bar has to be supported symmetrically
Therefore extension in each wire will be same
We know $\Delta x=\frac{F L}{A Y}$
Compare $1$ copper wire with another steel wire
$\frac{F L}{A_C Y_C}=\frac{F L}{A_S Y_S}$
$\Rightarrow \frac{A_S}{A_C}=\frac{Y_C}{Y_S}$ $\left\{\begin{array}{l}\text { Where, } \\ A_C-\text { Area of copper wire } \\ Y_C-\text { Young's modulus copper } \\ A_S-\text { Area of steel wire } \\ Y_S-\text { Young's modulus steel }\end{array}\right.$
Substitutuing value of $Y_C$ and $Y_s$
$\frac{\pi d_S^2}{4 \times \frac{\pi}{4} \times d_C^2}=\frac{110 \times 10^9}{190 \times 10^9}$
$\frac{d_S}{d_C}=\sqrt{\frac{11}{19}}$ $\left\{d_S \text {-diameter of steel wire } \ d_C \text {-diameter of copper wire }\right\}.$
$\frac{d_C}{d_S}=\sqrt{\frac{19}{11}}$
From free body diagram
$3 T=150 \,N$
$T=50 \,N$
Since the bar has to be supported symmetrically
Therefore extension in each wire will be same
We know $\Delta x=\frac{F L}{A Y}$
Compare $1$ copper wire with another steel wire
$\frac{F L}{A_C Y_C}=\frac{F L}{A_S Y_S}$
$\Rightarrow \frac{A_S}{A_C}=\frac{Y_C}{Y_S}$ $\left\{\begin{array}{l}\text { Where, } \\ A_C-\text { Area of copper wire } \\ Y_C-\text { Young's modulus copper } \\ A_S-\text { Area of steel wire } \\ Y_S-\text { Young's modulus steel }\end{array}\right.$
Substitutuing value of $Y_C$ and $Y_s$
$\frac{\pi d_S^2}{4 \times \frac{\pi}{4} \times d_C^2}=\frac{110 \times 10^9}{190 \times 10^9}$
$\frac{d_S}{d_C}=\sqrt{\frac{11}{19}}$ $\left\{d_S \text {-diameter of steel wire } \ d_C \text {-diameter of copper wire }\right\}.$
$\frac{d_C}{d_S}=\sqrt{\frac{19}{11}}$
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