A light rod of length $200 c m$ is suspended from the ceiling horizontally by means of two vertical wires of equal length tied to its ends. One of the wires is made of steel and is of cross-section $0.1 s q . c m$ and the other is of brass of cross-section $0.2 s q . c m$ . Find the position along the rod at which a weight may be hung to produce (i) equal stresses in both wires and (ii) equal strains in both wires. $Y_{b r a s s} = 10 \times 10^{11} d y n e / c m^{2}$ and $Y_{s t e e l} = 20 \times 10^{11} d y n e / c m^{2}$

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$200 c m; 100 c m$

$133.3 c m; 100 c m$

None

$167 c m; 50 c m$

answer is A.

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Detailed Solution

Stress same $F \propto A$

$\frac{F_{1}}{F_{2}} = \frac{A_{1}}{A_{2}} = \frac{1}{2} \Rightarrow F_{2} = 2 F_{1}$

$F_{1} x = F_{2} (2 - x)$

$F_{1} x = 2 F_{1} (2 - x) \Rightarrow x = 4 - 2 x \Rightarrow x = 1.333 m$

$x = 133.3 c m$

Strain same: $F \propto A y$

$\frac{F_{1}}{F_{2}} = \frac{A_{1}}{A_{2}} \frac{y_{1}}{y_{2}} = \frac{0.2}{0.1} \times \frac{10 \times 10^{11}}{20 \times 10^{11}} = 1$

$F_{1} x = F_{2} (2 - x) \Rightarrow x = 2 - x$

$x = 1 m = 100 c m$

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