Question

A 2$m$ long light metal rod $AB$ is suspended from the celling horizontallyby means of two vertical wires of equal length, tied to its ends. One wire is of brass and has cross-section of $0.2\times {10}^{-4}{m}^2$ and the other is of steel with $0.1\times {10}^{-4}{m}^2$ cross-section. In order to have equal stresser in ofthe two wires, a weight should be hung from the rod from end A at adistance of

• A. 66.6$cm$
• B. 133$cm$
• C. 44.4$cm$
• D. 155.6$cm$

Answer 1

The correct option is B 133$cm$

REF.Image

For stresses to b equal

$\frac{T_1}{A_1}=\frac{T_2}{A_2}\Rightarrow T_2=2T_1$

Then,

$\sum T=0$ about C

$T_{1}x=T_2(2-x)$

$\Rightarrow x=2(2-x)$

$x=4-2x$

$x=\frac{4}{3}m\approx 133cm$