⇦ | ![]() | ⇨ |
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 transistor-oscillator using a resonant circuit with an inductor L
- The magnetic flux linked with a circuit of resistance 100Ω increases from 10 to 60 Wb.
- An automobile engine develops 100 kW when rotating at a speed of 1800 rev/min
- If an A.C. mains supply is given to be 220 V, what would be the average e.m.f.
- A body of length 1 m having cross-sectional area 0.75 m² has heat flow through it
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