Question

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

• A. $0.2J$
• B. $10J$
• C. $20J$
• D. $0.1J$

Answer 1

Answer: (D)

$0.1J$

Elastic energy $=\frac{1}{2}\times F\times x$

$F=200N,x=1mm={10}^{-3}m$

$\therefore E=\frac{1}{2}\times 200\times 1\times {10}^{-3}=0.1J$