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Question

Two cubical blocks of side length a and 2a are stuck symmetrically as shown in the figure. The combined block is floating in water with the bigger block just submerged completely. The block is pushed down a little and released. Find the time period of its oscillations. Neglect viscosity.


A
π2ag
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B
2π2ag
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C
3π2ag
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D
2πag
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Solution

The correct option is C 3π2ag
In equilibrium, FB=W
ρω(2a)3=ρg[(2a)3+a3]
ρρω=89
When the block is depressed by x downwards from is equilibrium position, the excess buoyancy is the restoring force.
Restoring force =(a2x)ρωg
i.e ma=(a2x)ρωg
or (9a3)ρ.d2xdt2=(a2ρωg)x
d2xdt2=(g8a)x [ρρω=89]
ω1=g8a
Time for half oscillation t1=π8ag

When the block is raised above its mean position by distance x, the restoring force is
Frest=(4a2 ρωg)x
md2xdt2=(4a2 ρωg)x
9a3ρd2xdt2=4a2ρωgx
d2xdt2=(g2a)x
ω2=g2a
Time for half oscillation t2=π2ag
T=t1+t2=π2ag[2+1]=3π2ag is the time period of oscillation of block.

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