A cylindrical block of density rho is partial | vedclass

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(D) When the cylindrical block is in equilibrium,the buoyant force equals the weight of the block:$F_B = mg \Rightarrow (3\rho) A y g = \rho A h g$,wh

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$(A)$ and $(C)$ both

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(D) When the cylindrical block is in equilibrium,the buoyant force equals the weight of the block:$F_B = mg \Rightarrow (3ho) A y g = ho A h g$,where $y$ is the submerged depth and $h = 60\, cm$.$3y = h \Rightarrow y = h/3 = 60/3 = 20\, cm$.Since the block cannot be displaced upwards by more than its submerged depth without leaving the liquid,the maximum amplitude is $20\, cm$.For a small displacement $x$ from equilibrium,the restoring force is $F = -(	ext{change in buoyant force}) = -(A \cdot 3ho \cdot g) x$.The effective spring constant is $k = 3Aho g$.The mass of the block is $m = ho A h$.The time period $T$ is given by $T = 2\pi \sqrt{m/k} = 2\pi \sqrt{(ho A h) / (3Aho g)} = 2\pi \sqrt{h / (3g)}$.Substituting $h = 0.6\, m$ and $g = 9.8\, m/s^2$:$T = 2\pi \sqrt{0.6 / (3 	imes 9.8)} = 2\pi \sqrt{0.6 / 29.4} = 2\pi \sqrt{1/49} = 2\pi/7\, s$.

$A$ cylindrical block of density $\rho$ is partially immersed in a liquid of density $3\rho$. The plane surface of the block remains parallel to the surface of the liquid. The height of the block is $60\, cm$. The block performs $SHM$ when displaced from its mean position. [Use $g = 9.8\, m/s^2$]

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