A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N to the lower end. The weight stretches the wire by 1 mm . Then the elastic energy stored in the wire is

Question

A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N to the lower end. The weight stretches the wire by 1 mm . Then the elastic energy stored in the wire is (A) 0.2 J (B) 10 J (C) 20 J (D) 0.1 J

Tardigrade Admin · Accepted Answer

Elastic energy per unit volume = (1/2) × textstress × textstrain ∴ Elastic energy = (1/2) × textstress × textstrain × textvolume = (1/2)× ( textF/ textA)× (Δ textL/ textL)× ( textAL) =(1/2) textFΔ textL=(1/2)× text200× text10- 3 text=0.1 J

Q. A wire suspended vertically from one of its ends is stretched by attaching a weight of $200 N$ to the lower end. The weight stretches the wire by $1 mm$ . Then the elastic energy stored in the wire is

$0.2 J$

$10 J$

$20 J$

$0.1 J$

Elastic energy per unit volume = $\frac{1}{2} \times stress \times strain$
$∴$ Elastic energy = $\frac{1}{2} \times stress \times strain \times volume$
= $\frac{1}{2} \times \frac{F}{A} \times \frac{Δ L}{L} \times (AL)$
$= \frac{1}{2} F Δ L = \frac{1}{2} \times 200 \times 10^{- 3} =0.1 J$

Q. A wire suspended vertically from one of its ends is stretched by attaching a weight of 200N to the lower end. The weight stretches the wire by 1mm . Then the elastic energy stored in the wire is

0.2J

10J

20J

0.1J