A small ball of density $$4{{\rho }_{0}}$$ is released from rest just below the surface of a liquid. The density of liquid increases with depth as $$\rho ={{\rho }_{0}}(1+ay)$$ where $$a=2{{m}^{-1}}$$is a constant. Find the time period of its oscillation. (Neglect the viscosity effects).

Question

TuteeHUB · Accepted Answer

$$\frac{\pi }{\sqrt{5}}\sec $$

TuteeHUB · Answer

$$\frac{2\pi }{\sqrt{5}}\sec $$

TuteeHUB · Answer

$$\frac{\pi }{2\sqrt{5}}\sec $$

TuteeHUB · Answer

$$\frac{3\pi }{2\sqrt{5}}\sec $$

1.	A small ball of density \[4{{\rho }_{0}}\] is released from rest just below the surface of a liquid. The density of liquid increases with depth as \[\rho ={{\rho }_{0}}(1+ay)\] where \[a=2{{m}^{-1}}\]is a constant. Find the time period of its oscillation. (Neglect the viscosity effects).
A.	\[\frac{2\pi }{\sqrt{5}}\sec \]
B.	\[\frac{\pi }{\sqrt{5}}\sec \]
C.	\[\frac{\pi }{2\sqrt{5}}\sec \]
D.	\[\frac{3\pi }{2\sqrt{5}}\sec \]
Answer» B. \[\frac{\pi }{\sqrt{5}}\sec \]

A small ball of density \[4{{\rho }_{0}}\] is released from rest just below the surface of a liquid. The density of liquid increases with depth as \[\rho ={{\rho }_{0}}(1+ay)\] where \[a=2{{m}^{-1}}\]is a constant. Find the time period of its oscillation. (Neglect the viscosity effects).

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