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Question

A uniform cubical block of density ρb(in g/cc) and length a is released from rest from the position where one of its horizontal surface (lower) just touches the water (density ρw=1g/cc). The time period of oscillation of the block is (ρb=23 g/cc,neglect viscosity and surface tension)

A
2πpbapwg
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B
2πpwapbg
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C
4π3pbapwg+pbpwpb4pwa3pbg
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D
2π3pbapwg+pbpwpb2pwa3pbg
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Solution

The correct option is C 4π3pbapwg+pbpwpb4pwa3pbg
In case of equilibrium,

B=mg
ρ(y)A)g=ρbaAg
y0=ρbaρw

If we displace the body in downdwards direction, then
Frestoring=(Bmg)
Frestoring=(ρw(y0+y)AgρbAag)
Frestoring=ρwy0AgρwyAg+ρbAag

Substituting value of y0 we get,
Frestoring=ρbAagρwyAg+ρbAag
Frestoring=(ρwAg)y

Since restoring force is proportional to displacement from mean position, hence body is performing SHM motion.

So we can write,
Frestoring=mω2y=(ρwAg)y
ρbAaω2y=ρwAgy
ω=ρwgρba

Now since y0=2a3 ,
This situation can be better understood from the diagram below.

So amplitude can be calculated as
A=a2+a6=2a3

But by the time body comes down a3, it will be completely submerged, and from there on buoyancy force does not increase more. So the motion ceases to be SHM.

So phasor of the motion of body will be,
cosϕ=yA=a32a3=12
ϕ=π3

So body is performing SHM only for the phase of 2π2ϕ=4π3.
Hence time duration when body is performing SHM will be
T1=4π3ρbaρwg

Now for time taken for uniformly accelerated motion:

a=Bmgm
a=ρwρbρbg

Velocity when body has just reached its completely submerged position,
v=ωA2y2
v=ρwgρba4a29a29
v=ρwga3ρb

So time taken during uniformly accelerated motion,
T2=2va=2ρwa3ρbgρbρwρb

So total time =T1+T2=4π3pbapwg+pbpwpb4pwa3pbg


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