Q-A copper wire of length 2 2m and a steel wire of length 1 6m, both of diameter 3 0mm, - Physics - Mechanical Properties Of Solids

Question

Q-A copper wire
of length 2 2m and a steel wire of length 1 6m, both of diameter 3
0mm, are connected end to end When stretched by a load, the net
elongation is found to be 0 70mm Obtain the load applied

Vijay Yadav · Accepted Answer

Length of copper wire Lc= 2 2 m Length of steel wire Ls = 1 6 m Diameter d = 3 mm = 3 Ã— 10â€“3 m Cross sectional are = $π r^{2} = π (\frac{d}{2})^{2}$ $= 3 14 \times (\frac{3 \times 10^{- 3}}{2})^{2} = 3 14 \times 2 25 \times 10^{- 6} m^{2} = 7 1 \times 10^{- 6} m^{2}$ Net elongation (lc + ls) = 0 70 mm = 7 Ã— 10â€“4 m (1) Let load applied is F Ys = 2 0 Ã— 1011 N/m2 Tc = 1 3 Ã— 1011 N/m2 $Y_{s} = \frac{Stress}{Strain} = \frac{\frac{F}{A}}{(\frac{l_{s}}{L_{s}})} (2) Y_{c} = \frac{(\frac{F}{A})}{(\frac{l_{c}}{L_{c}})} (3)$ Apply (2) and (3) $\frac{Y_{s}}{Y_{c}} = \frac{(\frac{F}{A})}{(\frac{l_{s}}{L_{s}})} \times \frac{(\frac{l_{c}}{L_{c}})}{(\frac{F}{A})} = (\frac{L_{s}}{L_{c}}) * (\frac{l_{c}}{l_{s}})$ Plug in the values $\Rightarrow \frac{2 0 \times 10^{11}}{1 3 \times 10^{11}} = \frac{(2 2)}{(1 16)} \times (\frac{l_{c}}{l_{s}}) \Rightarrow \frac{l_{c}}{l_{s}} = \frac{2 \times 1 6}{1 3 \times 2 2} = 1 12 l_{c} = 1 12 l_{s} (4)$ From (1) and (4) $1 12 l_{s} + l_{s} = 7 \times 10^{- 4} m (2 12) l_{s} = 7 \times 10^{- 4} m l_{s} = 3 30 \times 10^{- 4} m$ From Equation (2) $Y_{s} = \frac{F \times L_{s}}{A \times l_{s}} \Rightarrow F = \frac{Y_{s} \times A \times l_{s}}{L_{s}} = \frac{2 \times 10^{11} \times 7 16 \times 10^{- 6} \times 3 30 \times 10^{- 4}}{1 6} = \frac{2 \times 7 1 \times 33}{1 6} = 292 87 N$

Q-A copper wire of length 2.2m and a steel wire of length 1.6m, both of diameter 3.0mm, are connected end to end. When stretched by a load, the net elongation is found to be 0.70mm.Obtain the load applied.