Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

The potential energy of a long spring when stretched by $2 \text{ cm}$ is $U$ . If the spring is stretched by $8 \text{ cm}$ , potential energy stored in it will be

$2 \U$

$4 \U$

$8 \U$

$16 \U$

Step-by-Step Solution

Potential energy of a spring $U = \frac{1}{2}kx^2$ . Thus $U \propto x^2$ . If $x$ increases from $2 \text{ cm}$ to $8 \text{ cm}$ (a factor of 4), the energy increases by a factor of $4^2 = 16$ . So, new energy is $16U$ .

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started