Q. The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8 cm the potential energy stored in it is :
A) 4U
B) U/8
C) 16U
D) U/4
Explanation
[Solution from source] Spring energy is U = ½kx², so U ∝ x². Increasing the stretch from 2 cm to 8 cm (4 times) multiplies the energy by 4² = 16, giving 16U. Assuming U ∝ x would give only 4U. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.