Physics Elasticity Question 51
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 1 mm. The energy stored in the wire due to stretching is
[MP PET 1989]
Options:
A) Zero
B) 0.05 joule
C) 100 joule
D) 500 joule
Show Answer
Answer:
Correct Answer: B
Solution:
$ W=\frac{1}{2}\times F\times l=\frac{1}{2}mgl $
$ =\frac{1}{2}\times 10\times 10\times 1\times {{10}^{-1}}=0.05\ J $
BETA
