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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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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A stone of mass 1kg is tied to a string 4m long and is rotated at constant speed
- When the wave of hydrogen atom comes from infinity into the first orbit,
- In a given reaction, ᴢXᴬ → ᴢ+1Yᴬ → ᴢ-1Kᴬ⁻⁴ → ᴢ-1Kᴬ⁻⁴
- A particle moves along a circle of radius 20/π m with constant tangential acceleration
- If an electron and a proton have the same de-Broglie wavelength, then the kinetic
Topics: Laws of Motion (103)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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