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Acceleration Due to Gravity question from NEET 2019

Q111NEET 2019Medium

Acceleration Due to Gravity

A mass falls from a height 'hh' and its time of fall 'tt' is recorded in terms of time period TT of a simple pendulum. On the surface of earth it is found that t=2Tt = 2T. The entire set up is taken on the surface of another planet whose mass is half of that of earth and radius the same. Same experiment is repeated and corresponding times noted as t′t' and T′T'.

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Answer: (a)

  1. Time of fall t=2hg∝1gt = \sqrt{\frac{2h}{g}} \propto \frac{1}{\sqrt{g}}.
  2. Pendulum period T=2πlg∝1gT = 2\pi\sqrt{\frac{l}{g}} \propto \frac{1}{\sqrt{g}}.
  3. So tT\frac{t}{T} does not depend on gg and stays 2.
  4. So the answer is (a) t′=2T′t' = 2T'.
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