A steel wire of length 3.2 m $$(Y_S = 2.0 \times 10^{11} \ N m^{-2})$$ and a copper wire of length 4.4 m $$(Y_C = 1.1 \times 10^{11} \ N m^{-2})$$, 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 $$\pi = \frac{22}{7}$$)

A steel wire (length 3.2 m, $$Y_S = 2.0 \times 10^{11}$$ N/m$$^2$$) and a copper wire (length 4.4 m, $$Y_C = 1.1 \times 10^{11}$$ N/m$$^2$$), both of radius 1.4 mm, are connected end to end. The net elongation is 1.4 mm. We wish to find the load required.

First, the radius of each wire is $$r = 1.4 \text{ mm} = 1.4 \times 10^{-3} \text{ m}$$. Hence, the cross-sectional area is $$A = \pi r^2 = \frac{22}{7} \times (1.4 \times 10^{-3})^2 = \frac{22}{7} \times 1.96 \times 10^{-6}$$ which simplifies to $$A = 22 \times 0.28 \times 10^{-6} = 6.16 \times 10^{-6} \text{ m}^2.$$

Since the wires are in series, the same load $$F$$ acts on both. Using the relation $$\Delta l = \frac{Fl}{YA}$$, we have $$\Delta l_S = \frac{F \times 3.2}{2.0 \times 10^{11} \times 6.16 \times 10^{-6}}$$ for the steel wire and $$\Delta l_C = \frac{F \times 4.4}{1.1 \times 10^{11} \times 6.16 \times 10^{-6}}$$ for the copper wire.

The total elongation is given by $$\Delta l_S + \Delta l_C = 1.4 \times 10^{-3} \text{ m},$$ so that $$F\!\left(\frac{3.2}{2.0 \times 10^{11} \times 6.16 \times 10^{-6}} + \frac{4.4}{1.1 \times 10^{11} \times 6.16 \times 10^{-6}}\right) = 1.4 \times 10^{-3}.$$

Substituting and simplifying each term yields $$\frac{3.2}{2.0 \times 10^{11} \times 6.16 \times 10^{-6}} = \frac{3.2}{1.232 \times 10^{6}} = 2.597 \times 10^{-6},$$ $$\frac{4.4}{1.1 \times 10^{11} \times 6.16 \times 10^{-6}} = \frac{4.4}{6.776 \times 10^{5}} = 6.494 \times 10^{-6},$$ and hence the sum is $$2.597 \times 10^{-6} + 6.494 \times 10^{-6} = 9.091 \times 10^{-6}.$$

From the above, we solve for $$F$$: $$F = \frac{1.4 \times 10^{-3}}{9.091 \times 10^{-6}} = \frac{1400}{9.091} \approx 154 \text{ N}.$$

Answer: Option D: 154