If the ratio of diameters, lengths and Young's modulus of the steel and copper wires shown in the fi

Question

If the ratio of diameters, lengths and Young's modulus of the steel and copper wires shown in the figure are $p, q$ and $s$ respectively, then the corresponding ratio of longitudinal strain developed in them would be

Professor PC Thomas Classes · Answer

(a) : Strain $\epsilon = \frac{\Delta L}{L} = \frac{	ext{Stress}}{Y} = \frac{F}{A Y} = \frac{4F}{\pi D^2 Y}$. Therefore, the ratio of strain in steel to copper is $\frac{\epsilon_S}{\epsilon_C} = \left(\frac{F_S}{F_C}ight) \left(\frac{D_C}{D_S}ight)^2 \left(\frac{Y_C}{Y_S}ight)$. The tension in the steel wire $F_S$ is due to the total mass $(5m + 2m)g = 7mg$. The tension in the copper wire $F_C$ is due to the lower mass $5mg$. We are given ratios $\frac{D_S}{D_C} = p$ and $\frac{Y_S}{Y_C} = s$. Thus, $\frac{\epsilon_S}{\epsilon_C} = \left(\frac{7mg}{5mg}ight) \left(\frac{1}{p}ight)^2 \left(\frac{1}{s}ight) = \frac{7}{5 p^2 s}$.