The submarine is another application of 'Archimedes Principle'. On surface, the submarine floats, with its conning tower and most of the deck, being clear of the water.
Submarine diving
The boat is provided with large ballast tanks which can be filled with water. This increases the weight of the submarine, so that it sinks into the sea, so that it sinks lower.
Horizontal rudders are titled to make the submarine drive downwards when the tanks are practically full and the boat is ready to submerge. When the submarine is ready to surface, the rudders are moved to drive the boat upwards, and compressed air is forced into the ballast tanks to drive the water out so that the submarine can rise.
Most of the marine animals also use this principle to remain at a selected level in the sea. For example, fish has an air sac, called a swim bladder in its body. This is filled with air and usually occupies about 5 per cent of its total body volume. Its size is adjusted so that the fish is posed at the depth at which is usually lives and feeds. At that level, the condition of its weight is exactly balanced by the upthrust it experiences. To swim up and down the fish uses fins.
Numericals :
1. A block wood of mass 24 kg floats in water. The volume of wood is 0.032 m3. Find
(i) the volume of the block below the surface of water
(ii) the density of wood (Take density of water 1000 kg m-3)
Suggested answer :
(i) Volume of the block below the surface of water = Volume of water displaced.
According to principle of floatation mass of displaced water
= mass of the block = 24 kg.
24 = volume of water displaced x density of water
24 = V' x 1000
= 0.024
Volume of the block below the surface of water = 0.024 m3
(ii) mass of block = mass of water displaced.
V x d1 = V'd2
= 7.5 x 102 kg m-3
2. A weather forecasting plastic balloon of volume 15 m3 contains hydrogen of density 0.09 kg m-3. The volume of the equipment carried by the balloon is negligible compared to its own volume. The mass of the empty balloon alone is 7.15 kg. The balloon is floating in air of density 1.3 kg m-3. Calculate:
(i) the mass of hydrogen in the balloon
(ii) the mass of hydrogen and the balloon
(iii) the total mass of the hydrogen, the balloon and the equipment if the mass of the equipment is 'x' kg.
(iv) the mass of air displaced by the balloon
(v) the mass of the equipment using the law of floatation.
Suggested answer :
(i) mass of hydrogen in the balloon = 15 x 0.09 = 1.35 Kg
(ii) mass of hydrogen + balloon = 1.35 + 7.15 = 8.50 kg
(iii) mass of hydrogen + balloon + equipment = (8.50 + x) kg
(iv) mass of air displaced = 15 x 1.3 = 19.5 kg
(v) Total mass of floating body = mass of air displaced.
8.5 + x = 19.5
x = 19.5 - 8.5
x = 11 kg.
