Question

A thick brass wire of length \(L\) and density \(\rho\) is suspended from a rigid support. Due to its own weight, \(\ell\) is the increase in length. Young’s modulus \(Y\) of brass wire in terms of density is ( \(g\) = acceleration due to gravity )

• A. \( Y = \dfrac{\rho g L}{4\ell} \)
• B. \( Y = \dfrac{\rho g L^2}{4\ell} \)
• C. \( Y = \dfrac{\rho g L}{3\ell} \)
• D. \( Y = \dfrac{\rho g L^2}{\ell} \)

Hint:

For a wire stretched by its own weight, extension is proportional to \(L^2\), not \(L\).

Answer 1

The Correct Option is D

Step 1: Consider a small element of the wire.
For a wire hanging under its own weight, the tension varies along the length. The extension due to self-weight is given by the standard result: \[ \ell = \frac{\rho g L^2}{Y}. \]
Step 2: Rearranging for Young’s modulus.
\[ Y = \frac{\rho g L^2}{\ell}. \]
Step 3: Conclusion.
Thus, the Young’s modulus of the brass wire is \[ Y = \dfrac{\rho g L^2}{\ell}. \]