A cylindrical object of outer diameter 20 cm and mass 2 kg floats in water with its axis vertical. If it is slightly depressed and then released, find the time period of the resulting simple harmonic motion of the object.

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Cross-sectional area of the cylinder =π(0.10)² m² 

If the cylinder is depressed to a depth of X meter, the net force on the cylinder upward =πX(0.10)² * 1000 kg-weight. 

=10πX kg-weight =10πgX Newton 

Mass of the cylinder =2 kg 

Acceleration =Force/mass =10πgX/2 =5πgX m/s² 

But acceleration is also =⍵²X, 

equating both we get, ⍵²X    =5πgX 

→⍵ =√(5πg) 

Hence the time period of the simple harmonic motion 

=2π/⍵ =2π/√(5πg)

 =2*3.14/√(5π*9.8)

 =0.50 s

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