Two springs with spring constants K₁=1500 N/m and K₂=3000 N/m

Two Springs With Spring Constants K1500 Nm And K3000 Nm Physics Question

Two springs with spring constants K₁=1500 N/m and K₂=3000 N/m are stretched by the same force. The ratio of potential energy stored in springs will be

Options

(a) 1:2
(b) 2:1
(c) 4:1
(d) 1:4

Correct Answer:

2:1

Explanation:

Force F=-Kxdisplacement (x),
Potential Energy = U =(1/2) Kx²
U = (1/2) K(F/K)² = (1/2)(F²/K)
U₁/U₂=(1/2)(F²/K₁)x(2K₂/F²)
K₂/K₁=3000/1500=2/1.

Related Questions:

  1. 300j work is done in sliding a 2kg block up an inclined plane of height 10m
  2. 180⁰ phase difference is obtained when light ray is reflected from
  3. Light propagates rectilinearly because of its
  4. Which one of the following statements is true, if half-life of a radioactive substance
  5. Which one of the following is not a unit of young’s modulus?

Topics: Laws of Motion (103)
Subject: Physics (2479)

Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score

18000+ students are using NEETLab to improve their score. What about you?

Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.

NEETLab Mobile App

Share this page with your friends

Be the first to comment

Leave a Reply

Your email address will not be published.


*