Question

A steel wire of length 3.2 m (Ys = 2.0 × 1011 Nm-2) and a copper wire of length 4.4 m (Yc = 1.1 × 1011 Nm-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:
$(Given: π = \frac{22}{7})$

• A. 360
• B. 160
• C. 1080
• D. 154

Answer 1

Answer: (D)

154

Answer 2

The correct answer is (D): 154
$Δl_s + Δl_c = 1.4$
$\frac{Wl_s}{Y_sA} + \frac{Wl_c}{Y_c×A} = 1.4×10^{-3}$
$W = \frac{1.4×10^{-3}}{[\frac{3.2}{2×(π×1.4×10^{-3})^2}+\frac{4.4}{1.1×(π×1.4×10^{-3})^2]}\frac{1}{10^{+11}}}$
$W≃154N$