Q. 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
A) Zero
B) $0.05\ joule$
C) $100\ joule$
D) $500\ joule$
Explanation
(b) $W = \frac{1}{2} \times F \times l = \frac{1}{2}mgl = \frac{1}{2} \times 10 \times 10 \times 1 \times 10^{- 3} = 0.05\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.