AP EAMCET · PHYSICS · Mechanical Properties of Solids
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 \(=0.7 \times 10^{11} \mathrm{Nm}^{-2}\) )
- A \(12 \pi \mathrm{N}\)
- B \(24 \pi \mathrm{N}\)
- C \(20 \pi \mathrm{N}\)
- D \(80 \pi \mathrm{N}\)
Answer & Solution
Correct Answer
(C) \(20 \pi \mathrm{N}\)
Step-by-step Solution
Detailed explanation
\(\Delta l=\frac{\mathrm{F} l_1}{\mathrm{y}_1 \mathrm{~A}}+\frac{\mathrm{F} l_2}{\mathrm{y}_2 \mathrm{~A}}\) \(\Rightarrow 0.6 \times 10^{-3}=\frac{\mathrm{F}}{\pi \times\left(10^{-3}\right)^2}\left[\frac{2.4}{1.2 \times 10^{11}}+\frac{0.7}{0.7 \times 10^{11}}\right]\)…
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