Wave Mechanics Question 529
Question: A 5 metre long wire is fixed to the ceiling. A weight of 10 kg is hung at the lower end and is 1 metre above the floor. The wire was elongated by t mm. The energy stored in the wire due to stretching is
Options:
A) Zero
B) 0.05 joule
C) 100 joule
D) 500 joule
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ W=\frac{1}{2}\times F\times l=\frac{1}{2}mgl $
$ =\frac{1}{2}\times 10\times 10\times 1\times {{10}^{-3}}=0.05J $
BETA
