A steel wire of length $3.2$ m $(Y_S = 2.0\times10^{11}\ \text{N m}^{-2})$ and a copper wire of length $4.4$ m $(Y_C = 1.1\times10^{11}\ \text{N m}^{-2})$, both of radius $1.4$ mm are connected end to end. When stretched by a load, the net elongation is found to be $1.4$ mm. The load applied, in Newton, will be: (Given $\pi = \dfrac{22}{7}$)
Answer: (D) $154$
The same tension $F$ acts in both wires:
$$\Delta L = \frac{F}{A}\left(\frac{L_S}{Y_S} + \frac{L_C}{Y_C}\right) = \frac{F}{A}\left(\frac{3.2}{2\times10^{11}} + \frac{4.4}{1.1\times10^{11}}\right) = \frac{F}{A}(5.6\times10^{-11})$$
$A = \dfrac{22}{7}(1.4\times10^{-3})^2 = 6.16\times10^{-6}\ \text{m}^2$.
$$F = \frac{1.4\times10^{-3}\times6.16\times10^{-6}}{5.6\times10^{-11}} = 154\ \text{N}$$
Solution by Sreeraj P, M.Sc Physics