Sol. When the box is immersed in the water, the air inside it will be compressed and water will enter the box. The volume of air in the box can be found from Boyle's Law:

h 0 SP 0 = h 1 SP 1 , .................. (1)

where, for the respective positions of the box (not immersed and immersed), h 0 and h 1 represent the height of the lid of the box above the level of water in it, P 0 and P 1 represent the pressure of air in the box and S is the area of the base. It is more convenient in the given instance to express pressure in terms of meters of the column of water. Substituting h 0 = 3, P 0 = 0.76 × 13.6 (meters of the + 3 – h 1 (meters of the column of water) P 1 = P 0 + 18.6 + (3 – h 1 ) = 10.3 + 18.6 + 3 – h 1 (meters of the column of water), we obtain from equation (1) a quadratic equation for h 1 :

h 1 = ;

solving this, we get h 1 = 1 m approx.

Thus the new (compressed) volume of air in the box equals approximately 1 m 3 . Neglecting the volume occupied by the sides of the box, we find that the upthrust equals the weight of water in a volume of 1 m 3 , i.e. 1 tonne wt.