Wave Mechanics Question 530
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 1 mm. Then the elastic energy stored in the wire is
Options:
A) 0.2 J
B) 10 J
C) 20 J
D) 0.1 J
Show Answer
Answer:
Correct Answer: D
Solution:
[d] 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 $
BETA
