A cubical box of wood of side 30 cm weighing 21.6 kg floats on water horizontally. The depth of immersion of box is
Question
A cubical box of wood of side 30 cm weighing 21.6 kg floats on water horizontally. The depth of immersion of box is
Solution
The box floats on water, which means the weight of the water displaced by the box is equal to the weight of the box itself. This is according to Archimedes' principle.
The weight of the box is given as 21.6 kg. The weight of water displaced by the box can be calculated using the formula:
Weight of water displaced = Volume of water displaced * Density of water * g
where g is the acceleration due to gravity (approximately 9.8 m/s^2) and the density of water is approximately 1000 kg/m^3.
The volume of water displaced by the box is equal to the volume of the part of the box that is submerged in the water. If we let h be the depth of immersion of the box, then the volume of the box submerged in the water is h * (side of the box)^2.
Setting the weight of the box equal to the weight of the water displaced, we get:
21.6 kg = h * (0.3 m)^2 * 1000 kg/m^3 * 9.8 m/s^2
Solving this equation for h, we get:
h = 21.6 kg / [(0.3 m)^2 * 1000 kg/m^3 * 9.8 m/s^2]
After calculating, we get:
h = 0.24 m or 24 cm
So, the depth of immersion of the box is 24 cm.
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