A cubical body (side 0.1 m and mass 0.02 kg) floats in water. It is pressed and then released so that it oscillates vertically. Find the time period of oscillation. (Density of water= $10^{3}$ $k g / m^{3}$ ).

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Solution

When a floating cubical body is displaced down through a small distance x, the excess buoyant force that is known as the restoring force is given as

$F = - V p g$
where v = volume of the excess liquid displaced = Ax
$F = - A p g x$ ; A = area of cross section of the cube.
-m $ω_{o s c}^{2} x = - A p g x$
$ω_{o s c}$ = $\sqrt{A p g / m}$
m = 0.02kg, p = $10^{3}$ kg/ $m^{3}$ , A = $(0.1)^{2}$ $m^{2}$

T = 2 $π$ $\sqrt{m / A p g} = 0.0885$ sec

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A cubical body (side 0.1 m and mass 0.02 kg) floats in water. It is pressed and then released so that it oscillates vertically. Find the time period of oscillation. (Density of water=103 kg/m3).

A cubical body (side 0.1 m and mass 0.02 kg) floats in water. It is pressed and then released so that it oscillates vertically. Find the time period of oscillation. (Density of water= $10^{3}$ $k g / m^{3}$ ).