Floatation- Long Answer Questions

Q1.  (i) State and explain the experiment for verification of Archimedes’ principle.

(ii) Name the devices based on Archimedes’ principle.

Ans. (i) Archimedes’ principle: The Archimedes’ principle states that if a body is immersed in a fluid, wholly or partly, then it loses its weight equal to the weight of fluid displaced by the immersed part of the body.
Experimental verification: Take a solid body which is heavier than water and also insoluble in water. Suspend it from the hook of a spring balance to find its weight in air. Let its weight in air be W1. Now take an overflow vessel filled with water up to the overflow mark and place a measuring cylinder below the overflow tube. Now suspend the given solid from hook to spring balance and lower the solid in the water till it is completely immersed in water. Note the reading of the spring balance. It gives the weight of solid body in water. Let it be W2. Therefore,  Loss of weight in solid when immersed in water = W1 –W2. The water which overflowed when solid was immersed is collected in the cylinder placed below the overflow tube. Measure this volume. Then weight of water displaced by solid = \(V\rho g \), where \(\rho \) = density of water and g= acceleration due to gravity.
It is found that W1 — W2 = \(V\rho g \),  i.e.. loss in weight of solid when immersed in water is equal to the weight of the water displaced by the body. This verifies Archimedes’ principle.                                                                

(ii) Devices based on Archimedes’ principle: Design of all those devices which float in the fluid is based on Archimedes’ principle. These devices such as hydrometers, lactometers, balloons, boats and ships, submarines, etc., work according to Archimedes’ principle.

Q2. (a) A cube of side 5 cm is immersed in water and then in saturated salt solution. In which will it experience a greater buoyant force? If each side of the cube is reduced to 4cm and then immersed in water, what will be the effect on the buoyant force experienced by the cube as compared to the first case for water. Give reason for each case.

(b) A ball weighing 4kg of density 4000 kg m³ is completely immersed in water of density 10 kg m^-3. Find the force of buoyancy on it. (Given g = 10 ms^-2.)

Ans. (a) The cube will experience a greater buoyant force in the saturated salt solution because density of the salt solution is greater than that of water.
The smaller cube will experience lesser buoyant force as its volume is lesser than the initial cube.
(b) Buoyant force = weight of the liquid displaced
                                = density of water X volume of water displaced x g
                                =\(1000\times \frac { 4 }{ 4000 } \times 10 \) =10 N.