Q. A light rod of length 100 cm is suspended from the ceiling horizontally by means of two vertical wires of equal lengths tied to the ends of the rod. One of the wires is made of steel and is of area of cross- section $ 0.1\text{ }c{{m}^{2}}. $ The other wire is of brass and of area of cross-section $ 0.2\text{ }c{{m}^{2}}. $ The position from the steel wire along the rod at which a load is to be placed to produce equal stresses in both wires is ( $ {{Y}_{steel}}=20\times {{10}^{11}} $ dyne $ c{{m}^{-2}};\text{ }({{Y}_{brass}}=10\times {{10}^{11}}dyne\text{ }c{{m}^{-2}}) $
EAMCETEAMCET 2009
Solution:
Given that, length of rod AB = 100 cm Taking moment about point O with $ {{T}_{1}} $ and $ {{T}_{2}} $ tension in steel and brass wires respectively. $ {{T}_{1}}(x)={{T}_{2}}(100-x) $ $ \frac{{{T}_{1}}}{{{T}_{2}}}=\frac{100-x}{x} $ ?(i) For equal stress in both the wires, $ \frac{{{T}_{1}}}{{{A}_{1}}}=\frac{{{T}_{2}}}{{{A}_{2}}} $ where $ {{A}_{1}} $ and $ {{A}_{2}} $ are cross-sectional areas of steel and brass wires respectively. or $ \frac{{{T}_{1}}}{{{T}_{2}}}=\frac{{{A}_{1}}}{{{A}_{2}}} $ $ \therefore $ $ \frac{{{T}_{1}}}{{{T}_{2}}}=\frac{0.1}{0.2}=\frac{1}{2} $ ?(ii) From Eq. (ii), substituting the value of $ \frac{{{T}_{1}}}{{{T}_{2}}} $ in Eq. (i). $ \therefore $ $ \frac{1}{2}=\frac{100-x}{x} $ or $ x=200-2x $ or $ x=\frac{200}{3}\,cm $
