Q180 / 225 · JEE Main 2003 · Physics
A wire suspended vertically from one of its ends is stretched by attaching a weight of $200 \mathrm{~N}$ to the lower end. The weight stretches the wire by $1 \mathrm{~mm}$. Then the elastic energy stored in the wire is
- A$0.2 \mathrm{~J}$
- B$10 \mathrm{~J}$
- C$20 \mathrm{~J}$
- Correct answerD$0.1 \mathrm{~J}$
Solution
Elastic energy $=\frac{1}{2} \times \mathrm{F} \times \mathrm{x}$
$$
\mathrm{F}=200 \mathrm{~N}, \mathrm{x}=1 \mathrm{~mm}=10^{-3} \mathrm{~m} \quad \therefore \mathrm{E}=\frac{1}{2} \times 200 \times 1 \times 10^{-3}=0.1 \mathrm{~J}
$$
Hence, the correct answer is option: 'D'.