JEE Main 2015PhysicsOscillationsHardMCQ

JEE Main 2015 — Oscillations Question with Solution

JEE Main 2015 (11 Apr Online)

Question

A cylindrical block of wood (density = 650 kg m-3), of base area 30 cm2 and height 54 cm, floats in a liquid of density 900 kg m-3. The block is depressed slightly and then released. The time period of the resulting oscillations of the block would be equal to that of a simple pendulum of length (nearly) :

Choose an option

Show full solutionCorrect option: C
Correct answer
C39 cm

Step-by-step explanation

Case-I

In equilibrium

Fup=mg

ViρLg=mg

AhρLg=ALρsg………1



Case-II

Let block is pressed by x more in liquid

Fnet=Fup1-mg

=Vi1ρLg-mg

Fnet=A h+xρLg-ALρsg

Fnet=AhρLg+AxρLg-ALρsg. . . . . . . . . . (2)

from equation (1) & (2)

Fnet=AxρLg restoring force in upward direction

α=Fnetm=AρLgAρsLx

α=ω2x for SHM

and ω2=gl

so gl=ρLgρsL

so l=ρsLρL=650×54900=39cm

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About this question

This is a previous-year question from JEE Main 2015, 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.