| ⇦ | 
| ⇨ |
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: - At 0⁰C, 15 gm of ice melts to form water at 0⁰C. The change in entropy is
- A neutron is moving with a velocity u.It collides head on and elastically with an atom
- A body under the action of a force F⃗ = 6i⃗–8j⃗+10k⃗, acquires an accelerationof 1 m/s². The mass of this body must be
- A spring of spring constant 5 x 10³ Nm⁻¹ is streched intially by 5 cm from
- The unit of electric permittivity is
Topics: Laws of Motion
(103)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- At 0⁰C, 15 gm of ice melts to form water at 0⁰C. The change in entropy is
- A neutron is moving with a velocity u.It collides head on and elastically with an atom
- A body under the action of a force F⃗ = 6i⃗–8j⃗+10k⃗, acquires an accelerationof 1 m/s². The mass of this body must be
- A spring of spring constant 5 x 10³ Nm⁻¹ is streched intially by 5 cm from
- The unit of electric permittivity is
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