Physics · Work, Power & Energy

NEET (UG) 2023 — Question 16

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

Option A:
4 U
Option B:
8 U
Option C:
16 U
Correct
Option D:
2 U

Answer: C

Step-by-step solution

Potential energy stored in spring $U=\frac{1}{2} K x^2$

\begin{aligned} & U=\frac{1}{2} K(2)^2 \text { where } x=2 \mathrm{~cm} \\ & U=\frac{1}{2}(K) \cdot(4) \\ & U=2 K \\ & U^{\prime}=\frac{1}{2} K(8)^2 \\ & U^{\prime}=\frac{1}{2} K \times 64=32 K \end{aligned}

On dividing (i) by (ii)

\begin{aligned} & \frac{U}{U^{\prime}}=\frac{2 K}{32 K}=\frac{1}{16} \\ & U^{\prime}=16 U \end{aligned}

Answer key and solution verified before publishing.

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Exam: NEET (UG) 2023
Paper: 2023 paper
Subject: Physics
Chapter: Work, Power & Energy
Topic: Kinetic Energy and Work-Energy Theorem

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