Wave Mechanics Question 520
Question: When a force is applied on a wire of uniform cross-section area $ 3\times {{10}^{-6}}m^{2} $ and length 4m, the increase in length is 1 mm. Energy stored in it will be ( $ Y=2\times 10^{11}N/m^{2} $ )
Options:
A) 6250J
B) 0.177J
C) 0.075 J
D) 0.150J
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ e=\frac{\Delta Y}{\ell }=\frac{1\times {{10}^{-3}}}{4} $
$ u=\frac{e^{2}y}{2}={{( \frac{{{10}^{-3}}}{4} )}^{2}}\times \frac{2\times 10^{11}}{2}=0.075J $
BETA
