⇦ | ⇨ |
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: - A coin placed on a rotating turn table just slips,if it is placed at a distance
- Find the maximum velocity with which a train can be moved on a circular track
- Dimensional formula for electrical resistance R is given by
- The density of ice is 0.9 g/cc and that of sea water is 1.1 g/cc
- On bombarding U²³⁵ by slow neutron, 200 MeV energy is released. If the power output
Topics: Laws of Motion
(103)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A coin placed on a rotating turn table just slips,if it is placed at a distance
- Find the maximum velocity with which a train can be moved on a circular track
- Dimensional formula for electrical resistance R is given by
- The density of ice is 0.9 g/cc and that of sea water is 1.1 g/cc
- On bombarding U²³⁵ by slow neutron, 200 MeV energy is released. If the power output
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.
Share this page with your friends
Leave a Reply