NEET 2019 Odisha · PhysicsSurface gPrevious Year Question
A mass falls from height h and its fall-time t is measured in units of the period T of a simple pendulum. On earth t = 2T. The setup is taken to another planet of half the earth's mass and the same radius, giving times t' and T'. Then:
Answer: (A) t' = 2T'. Correct Answer (A): t' = 2T' Solution: Free-fall time t = √(2h/g) ∝ 1/√g and pendulum period T = 2π√(L/g) ∝ 1/√g.
- A.t' = 2T'✓
- B.t' > 2T'
- C.t' < 2T'
- D.cannot be determined
Correct Answer
(A) t' = 2T'
Solution & Explanation
Correct Answer (A): t' = 2T' Solution: Free-fall time t = √(2h/g) ∝ 1/√g and pendulum period T = 2π√(L/g) ∝ 1/√g. The ratio t/T is independent of g, so t' = 2T' on any planet.
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