Different forces are acting on the wire which is hanging. Because different part of wire will be acted upon by different part of wire will be acted upon by different forces.
Therefore,
Consider a small element dx at a distance x from the load (x = 0). Let T(x) and T(x + dx) are tensions on the two cross sections a distance dx apart. Then
T(x + dx) + T(x) = dmg = μgdx (where μ is the mass/length) [∵ dm = μdx].
\(\frac{dT}{dx}\)dx = μgdx
or dT = μgdx [∵ dT = T(x + dx) + T(x)].
or T(x) = μgx + C
At x = 0, T(0) = Mg or C = Mg
∴ T(x) = μgx + Mg
Let the length dx at x increase by dr. then Young’s Modulus, Y = \(\frac{Stress}{Strain}\)
\(\frac{\frac{T(x)}{A}}{\frac{dr}{dx}}\) = Y
Or \(\frac{dr}{dx}\) = \(\frac{1}{YA}\)T(x)
Or dr = \(\frac{1}{YA}\) \(∫^L_0\) (μgx + Mg)dx
Or \(\frac{1}{YA}\)[\(\frac{μgx^2}{\text w}\) + Mgx\(]^L_0\)
Or \(\frac{1}{YA}\)\(\bigg[\frac{MgL^2}{2}+Mgl\bigg]\) [∵ μL = m]
(m is the mass of the wire)
A = π × (10-3)2m2,
Y = 200 × 109 Nm-2
m = π × (10-3)2 × 10 × 7860 kg
∴ r = \(\frac{1}{2×10^{11}×π×10^{−6}}\) \(\bigg[\frac{π×786×10^{−3}×10×10}{2} + 25 × 10 × 10\bigg]\) = [196.5 × 10−6 + 3.98 × 10−3] ≈ 4 × 10−3m
(b) Clearly, the maximum tension would be at x = L
T = μgL + Mg = (m + M)g [∵ m = μL]
The yield force = (yield strength, y) area = 250 × πN
At yield point, T = yield Force
(M + M)g = 250 × π
m = π × (10-3)2 × 10 × 7860 < < M,
∴ Mg ≈ 250 × π
Hence, M = \(\frac{250 ×π}{10}\) = 25 × π ≈ 75kg.
