Question

AB is an iron wire and CD is a copper wire of same length and same cross-section. BD is a rod of length 0.8 m. A load G=2kg-wt is suspended from the rod. At what distance x from point B should the load be suspended for the rod to remain in a horizontal position $(Y_{Cu}=11.8\times {10}^{10}N/m^2,Y_{Fe}=19.6\times {10}^{10}N/m^2)$

• A. 0.1 m
• B. 0.3 m
• C. 0.5m
• D. 0.7 m

Answer 1

The correct option is B 0.3 m

$\text{Let}E\text{be the point of Suspension}$

$\begin{array}{c}BD=0.8m\\ \Delta l_1=\Delta l_2\text{(elongation of}AB=\text{elengation of}CD)\\ \frac{F_{1}l}{A{Y}_1}=\frac{F_2\ell }{A{Y}_2}\\ \frac{T_1}{T_2}=\frac{Y_1}{Y_2}\\ T_1/T_2=\frac{Y_1}{Y_2}=\frac{19.6\times {10}^{10}}{11.8\times {10}^{10}}=\frac{9.8}{5.9}\end{array}$

$\begin{array}{c}\text{Applying torque equation at}E\\ \begin{array}{c}{T}_{1}x=T_2(0.8-x)\\ \frac{T_1}{T_2}=\frac{0.8}{x}-1\\ x=\frac{0.8}{9.8+5.9}\times 5.9\\ \Rightarrow x=0.3m\end{array}\end{array}$