Question

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

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

Answer 1

The correct option is A $0.1$

For equilibrium of the system

$\Rightarrow F=200=k\left({10}^{-3}\right)$

$\Rightarrow k=2\times {10}^5$

Energy stored in wire $=0.5\times k{x}^2=0.5\times 2\times {10}^5\times ({10}^{-3}{)}^2=0.1J$

Hence correct answer is option $A$