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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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Subject: Physics
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Find out the e.m.f. produced when the current changes from 0 to 1 A in 10 sec.
- An object is seen through a simple microscope of focal length 12 cm.
- Electrons of mass m with de-Broglie wavelength λ fall on the target in an X-ray tube.
- If ₉₂U²³⁸ undergoes successively 8 ∝-decays and 6 β-decays, the resulting nucleus is
- The quantity having the same units in all systems of unit 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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