Question

A wire of length L, cross-sectional area A, density \( \rho \) and Young's modulus Y is suspended vertically. The elastic energy stored in the wire due to its own weight is \( \frac{2 p^{2} g^{2} A L^{2}}{3 Y} \) \( \frac{\rho^{2} g^{2} A^{2} L^{2}}{12 Y} \) \( \frac{\rho^{2} g^{2} A L^{3}}{6 Y} \) \( \frac{\rho^{2} g A L^{2}}{6 Y} \)

Answer 1

Area = A 

Density = ρ

Length = L

Yang modulus = Y

Mass per unit length = M/L

Mass of AB = \((\frac{M}L)x = \frac{Mx}L\)

\(\therefore\) M = ρ.v = ρ(A .L)

Mass of AB = \(\frac{\rho ALx}L\)

= ρ Ax

T = mg

T = ρAxg

Stress =  ρAx

Streacting energy stored in wire

E = \(\frac{1}2\) x stress x strain x volume

\(\because\) Strain = \(\frac{stress}y\)

then

E = \(\frac{1}2\) x \(\frac{stress}y\) x stress x volume

E = \(\frac{1}{2y}\)(stress)² x volume

\(\because\) stress = ρgx

for dx

dE = \(\frac{1}{2y}\)(ρgx)² x (dx) x A

∫dE = \(\int\limits_0^L\frac12\times\frac{\rho^2g^2x^2}{y}\)A dx

∫dE = \(\frac{\rho^2g^2A}{2y}(\frac{x^3}3)_0^L\)

E = \(\frac{\rho^2Ag^2L^3}{6y}\)