JEE Main 2014PhysicsOscillationsHardMCQ

JEE Main 2014 — Oscillations Question with Solution

JEE Main 2014 (09 Apr Online)

Question

The amplitude of a simple pendulum, oscillating in air with a small spherical bob, decreases from 10 cm to 8 cm in 40 seconds. Assuming that Stokes law is valid, and ratio of the coefficient of viscosity of air to that of carbon dioxide is 1.3, the time in which amplitude of this pendulum will reduce from 10 cm to 5 cm in carbondioxide will be close to (ln 5 = 1.601, ln 2 = 0.693).

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Show full solutionCorrect option: D
Correct answer
D161 s

Step-by-step explanation

A= A 0 e −bt 2m

b is related to viscosity

8=10 e − b a 40 2m

5=10 e − b C o 2 t 2m

ln 5 4 ln2 = 40 t ×1.3

t= 40×( ln2 )×1.3 ln5−2ln2

= 40×0.693×1.3 1.601−2×0.693

≃161sec

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About this question

This is a previous-year question from JEE Main 2014, covering the Oscillations chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.