Given :
Y = 12.5 × 1011 dyne cm-2
= 12.5 × 1010 Nm-2
Diameter, D = 2.5 mm = 2.5 × 10-3 m.
∴ radius, r = \(\frac{D}{2}\) = 1.25 × 10-3 m.
F = 100 kgf = 100 × 9.8 N = 980 N. \(\frac{∆L}{L}\) × 100 = ?
A = πr2 = π(1.25 × 10-3)2 m2.
From the relation, Y = \(\frac{FL}{A∆L},\)
We get % increase in length = \(\frac{∆L}{L}\) × 100
= \(\frac{F}{AY}\) × 100 = \(\frac{F}{πr^2Y}\) × 100
= \(\frac{980}{3.142×(1.25×10^−3)×12.5×10^{10}}\) × 100
= 0.16 %