A uniform wire of length $L$, density $\rho$ and Young's modulus $Y$ hangs vertically from a rigid support. The extension of the wire due to its own weight is
Answer: (D) $\dfrac{\rho g L^2}{2Y}$
Concept. The tension in a hanging wire varies along it: a point at height $x$ above the free end supports only the wire below it. Add the stretches of thin slices.
Step 1. Tension at height $x$. The part below weighs $\rho Axg$, so the tension there is $T(x) = \rho Agx$.
Step 2. Stretch of a slice $dx$.
$$d(\Delta L) = \frac{T(x)\,dx}{AY} = \frac{\rho gx}{Y}dx$$
Step 3. Integrate.
$$\Delta L = \int_0^L \frac{\rho gx}{Y}dx = \frac{\rho gL^2}{2Y}$$
(The same as the whole weight acting through half the length.)
Answer. $\dfrac{\rho gL^2}{2Y}$ (Option D).
Solution by Sreeraj P, M.Sc Physics