A copper wire of length $2.4 \mathrm{~m}$ and an aluminum wire of length $0.7 \mathrm{~m}$, both having…

A copper wire of length $2.4 \mathrm{~m}$ and an aluminum wire of length $0.7 \mathrm{~m}$, both having diameter $2 \mathrm{~mm}$, are connected end to end. When stretched by a load, the obtained elongation is found to be $0.6 \mathrm{~mm}$.The applied load is (Young's modulus of copper $=1.2 \times 10^{11} \mathrm{Nm}^{-2}$ and Young's modulus of aluminum $=0.7 \times 10^{11} \mathrm{Nm}^{-2}$ )
  1. $12 \pi \mathrm{N}$
  2. $24 \pi \mathrm{N}$
  3. $20 \pi \mathrm{N}$
  4. $80 \pi \mathrm{N}$

Solution

$\Delta l=\frac{\mathrm{F} l_1}{\mathrm{y}_1 \mathrm{~A}}+\frac{\mathrm{F} l_2}{\mathrm{y}_2 \mathrm{~A}}$ $\Rightarrow 0.6 \times 10^{-3}=\frac{\mathrm{F}}{\pi \times\left(10^{-3}\right)^2}\left[\frac{2.4}{1.2 \times 10^{11}}+\frac{0.7}{0.7 \times 10^{11}}\right]$ $\begin{aligned} & \Rightarrow 0.6 \pi \times 10^{-9}=\mathrm{F}\left[2 \times 10^{-11}+10^{-11}\right] \\ & \Rightarrow 0.6 \pi \times 10^{-9}=\mathrm{F} \times 3 \times 10^{-11} \\ & \Rightarrow \quad \mathrm{F}=\frac{0.6 \pi \times 10^{-9}}{3 \times 10^{-11}}=20 \pi \mathrm{N}\end{aligned}$

Asked in: AP EAMCET 2022 (05 Jul Shift 2)

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