A steel wire of length 2 m is stretched through 2 mm. What is elastic potential energy stored in a wire of cross-sectional area 4 mm 2 in stretched condition (Y=2 × 1011 Nm -2)

Question

A steel wire of length 2 m is stretched through 2 mm. What is elastic potential energy stored in a wire of cross-sectional area 4 mm 2 in stretched condition (Y=2 × 1011 Nm -2) (A) 0.8 J (B) 1.2 J (C) 2.4 J (D) 1.6 J

Tardigrade Admin · Accepted Answer

(Δ l/l)=(2/2) × 10-3=10-3 Stress =Y × strain =2 × 1011 × 10-3 =2 × 108 Nm 2 Volume =4 × 10-6 × 2=8 × 10-6 m 3 Energy stored =(1/2) × stress × strain × volume =(1/2) × 2 × 108 × 10-3 × 8 × 10-6=0.8 J

Q. A steel wire of length 2m is stretched through 2mm. What is elastic potential energy stored in a wire of cross-sectional area 4mm2 in stretched condition (Y=2×1011Nm−2)

0.8 J

1.2 J

2.4 J

1.6 J

lΔl​=22​×10−3=10−3 Stress =Y× strain =2×1011×10−3 =2×108Nm2 Volume =4×10−6×2=8×10−6m3 Energy stored =21​× stress × strain × volume =21​×2×108×10−3×8×10−6=0.8J

Q. A steel wire of length $2 m$ is stretched through $2 mm$ . What is elastic potential energy stored in a wire of cross-sectional area $4 m m^{2}$ in stretched condition $(Y = 2 \times 1 0^{11} N m^{- 2})$