KCET · Physics · Mechanical Properties of Solids
A thick metal wire of density \(\rho\) and length \(L\) is hung from a rigid support. The increase in length of the wire due to its own weight is ( \(Y=\) Young's modulus of the material of the wire)
- A \(\frac{\partial g L}{Y}\)
- B \(\frac{1}{2} \frac{\rho g L^2}{Y}\)
- C \(\frac{\rho g L^2}{Y}\)
- D \(\frac{1}{4 Y} \rho g L^2\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{2} \frac{\rho g L^2}{Y}\)
Step-by-step Solution
Detailed explanation
Weight of the metal wire, \(w=m g\)
\(\therefore\) Longitudinal stress \(=\frac{W}{A}=\frac{m g}{A}\)
\(\therefore \quad Y=\frac{\text { Longitudinal stress }}{\text { Longitudinal strain }}=\frac{\frac{m g}{A}}{\frac{\Delta l}{\left(\frac{l}{2}\right)}}\)
\(\left[\because\right.\) Weight of the wire acts at its centre \(\left.\therefore L^{\prime}=\frac{L}{2}\right]\)
\(\Rightarrow \quad \Delta l=\frac{m g l}{2 Y A}\)
\(=\frac{m}{A L} \cdot \frac{L^2 g}{2 Y}\)
\(=\frac{\rho L^2 g}{2 Y}\)
\(\therefore\) Longitudinal stress \(=\frac{W}{A}=\frac{m g}{A}\)
\(\therefore \quad Y=\frac{\text { Longitudinal stress }}{\text { Longitudinal strain }}=\frac{\frac{m g}{A}}{\frac{\Delta l}{\left(\frac{l}{2}\right)}}\)
\(\left[\because\right.\) Weight of the wire acts at its centre \(\left.\therefore L^{\prime}=\frac{L}{2}\right]\)
\(\Rightarrow \quad \Delta l=\frac{m g l}{2 Y A}\)
\(=\frac{m}{A L} \cdot \frac{L^2 g}{2 Y}\)
\(=\frac{\rho L^2 g}{2 Y}\)
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