The elastic potential energy stored in a steel wire of length 20 m stretched through 2 cm is 80 J. The cross sectional area of the wire is __________ mm2. (Given, y=2.0×1011Nm–2) [2023]
(40)
Energy per unit volume = 12 stress × strain
Energy = 12 stress× strain × volume
80=12×Y×strain2A×l
80=12×2×1011×(2×10–2)240×A×20
20=10+720×A
40×10–6m2=A
A=40 mm2
