A mass falls from a height ' h ' and its time of fall ' t ' is recorded in terms of time period T of a simple pendulum. On the surface of earth it is…
NEET UG 2019 — Physics Waves & Oscillations
2019mcqmedium
A mass falls from a height ' h ' and its time of fall ' t ' is recorded in terms of time period T of a simple pendulum. On the surface of earth it is found that t=2T. The entire set u is taken on the surface of another planet whose mass is half of earth and radius the same. Same experiment is repeated and corresponding times noted as t′ and T′.
Official previous-year question
Held on 30 Apr 2019 · Verified 9 Jul 2026.
Options
A
t′=2T′
B
t′>2T′
C
t′<2T′
D
t′=2T′
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Solution
The distance covered by the mass falling from height ' h ' during its time of fall ' t ' is given by s=h=ut+21gt2 As, u=0⇒h=21gt2⇒t=g2h The time period of simple pendulum is T=2πgl where, lis the length of the pendulum. From Eq. (i) and (ii), since ' h ' and ' l ' are constant so, we can conclude that, t∝gl and T∝gl∴Tt=l Thus, the ratio of time of fall and time period of pendulum is independent of value of gravity (g) or any other parameter like mass and radius of the planet. Thus, the relation between t′ and T′ on another planet irrespective of its mass or radius will remains same as it was on earth i.e. t′=2T′
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