A particle is dropped vertically from rest from a height. The time taken by it to fall…
Physics · JEE Advanced · NTA Exams — Kinematics
A particle is dropped vertically from rest from a height. The time taken by it to fall through successive distances of 1 m each will then be
- All equal, being equal to \(\sqrt{2 / g}\) second
- In the ratio of the square roots of the integers i.e, \(1: \sqrt{2}: \sqrt{3} \ldots\)
- In the ratio of the difference in the square roots of the integers i.e. \(\sqrt{1}:(\sqrt{2}-\sqrt{1}):(\sqrt{3}-\sqrt{2}):(\sqrt{4}-\sqrt{3}) \ldots\)
- In the ratio of the reciprocal of the square roots of the integers i.e.,\(\frac{1}{\sqrt{1}}: \frac{1}{\sqrt{2}}: \frac{1}{\sqrt{3}}: \frac{1}{\sqrt{4}}\)
Answer
(C) In the ratio of the difference in the square roots of the integers i.e. 1 :( 2 - 1 ):( 3 - 2 ):( 4 - 3 )
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