JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Elasticity & SHM

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 ______ mm

^{2}

.
(Given,

Y = 2.0 \times 1 0^{11}

^{- 2}

)

Answer: 40

Solution

The elastic potential energy per unit volume is

\frac{1}{2} \times stress \times strain = \frac{1}{2} Y (strain)^{2}

.

Total energy:

U = \frac{1}{2} Y \times (strain)^{2} \times A \times ℓ

where strain

= \frac{Δ ℓ}{ℓ} = \frac{0.02}{20} = 1 0^{- 3}

80 = \frac{1}{2} \times 2 \times 1 0^{11} \times (1 0^{- 3})^{2} \times A \times 20

80 = 1 0^{11} \times 1 0^{- 6} \times 20 \times A

80 = 2 \times 1 0^{6} \times A

A = 40 \times 1 0^{- 6} m^{2} = 40 mm^{2}

JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Integer answer