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JEE Main 201511 AprilOnline
Exam
JEE Main
Year
2015
Date
11 April
Mode
Online
Subject
Physics
Chapter
Oscillations

JEE Main 2015 · 11 April · Online · Physics · Oscillations

A cylindrical block of wood (density = 650kg m ^ - 3 ), of base...

A cylindrical block of wood (density = 650kgm−3650kg{m}^{- 3}), of base area 30cm230{cm}^{2} and height 54cm54cm, floats in a liquid of density 900kgm−3900kg{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) :

A cylindrical block of wood (density = 650kg m ^ - 3 ), of base area 30 cm ^ 2 and height 54cm , floats in a liquid of

Options

  1. A52cm52cm
  2. B26cm26cm
  3. C39cm39cm
  4. D65cm65cm

Correct answer

C. 39cm39cm

Step-by-step solution

Case-I

In equilibrium

Fup=mg{F}_{up} = mg

ViρLg=mg{V}_{i}{ρ}_{L}g = mg

AhρLg=ALρsg………(1)Ah{ρ}_{L}g = AL{ρ}_{s}g \ldots \ldots \ldots \left(1\right)

A cylindrical block of wood (density = 650kg m ^ - 3 ), of base area 30 cm ^ 2 and height 54cm , floats in a liquid of

Case-II

Let block is pressed by x more in liquid

Fnet=Fup1−mg{F}_{net} = {F}_{up}^{1} - mg

=Vi1ρLg−mg= {V}_{i}^{1}{ρ}_{L}g - mg

Fnet=A(h+x)ρLg−ALρsg{F}_{net} = A\left(h + x\right){ρ}_{L}g - AL{ρ}_{s}g

Fnet=AhρLg+AxρLg−ALρsg{F}_{net} = Ah{ρ}_{L}g + Ax{ρ}_{L}g - AL{ρ}_{s}g. . . . . . . . . . (2)

from equation (1) & (2)

Fnet=AxρLg{F}_{net} = Ax{ρ}_{L}grestoring force in upward direction

α=Fnetm=(AρLgAρsL)x\alpha = \dfrac{{F}_{net}}{m} = \left(\dfrac{A{ρ}_{L}g}{A{ρ}_{s}L}\right)x

α=ω2xforSHM\alpha = {\omega}^{2}x for SHM

and ω2=gl{\omega}^{2} = \dfrac{g}{l}

so gl=ρLgρsL\dfrac{g}{l} = \dfrac{{ρ}_{L}g}{{ρ}_{s}L}

so l=ρsLρL=650×54900=39cml = \dfrac{{ρ}_{s}L}{{ρ}_{L}} = \dfrac{650 \times 54}{900} = 39cm
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