A barrel that is 0.25 m in radius and 1.0 m in height is filled with oil (density of 920 kg/m³) and dropped into a lake (density of 1000 kg/m³). The axis of the barrel is perpendicular to the surface of the water. If the barrel is displaced from equilibrium by a small amount and released, the resulting period of oscillation is A) 0.50 s. B) 1.9 s. C) 1.5 s. D) 0.67 s.

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Quizplus · Accepted Answer

The period of oscillation can be determined using the formula for the period of a floating object: $T = 2\pi\sqrt{\frac{m}{\rho_{water}gA}}$, where $m$ is the mass of the barrel, $\rho_{water}$ is the density of water, $g$ is the acceleration due to gravity, and $A$ is the cross-sectional area of the barrel that is perpendicular to the direction of oscillation. Given the radius $r = 0.25$ m, the height $h = 1.0$ m, and the density of oil $\rho_{oil} = 920$ kg/m³, we first find the mass of the oil-filled barrel as $m = \rho_{oil} \times \text{volume of the barrel} = 920 \times \pi \times r^2 \times h$. The cross-sectional area $A = \pi r^2$. Substituting these values into the formula and using $\rho_{water} = 1000$ kg/m³ and $g = 9.8$ m/s², we find that the period $T$ closely matches the value given in option B, 1.9 seconds, after performing the calculation.

A Barrel That Is 0