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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.
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Topics: Laws of Motion
(103)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- If voltage across a bulb rated 220 volt-100 Watt drops by 2.5% of its rated value
- In stretching a spring by 2 cm, energy stored is given by U. Had is been stretched
- If the pressure of gas is increased from 1.01×10⁵ Pa to 1.165×10⁵ Pa and volume
- At 10⁰C the value of the density of a fixed mass of an ideal gas divided by its pressure
- The minimum number of NAND gates used to construct an OR gate is
Topics: Laws of Motion (103)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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