Shaskolskaya & Eltsin Problem 157

Physics

Problem

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: h0SP0=h1SP1h_0 SP_0 = h_1 SP_1, (1) where, for the respective positions of the box (not immersed and immersed), h0h_0 and h1h_1 represent the height of the lid of the box above the level of water in it, P0P_0 and P1P_1 represent the pressure of air in the box and SS is the area of the base. It is more convenient in the given instance to express pressure in terms of metres of the column of water. Substituting h0=3h_0 = 3, P0=0.76×13.6P_0 = 0.76 \times 13.6 (metres of the column of water), P1=P0+18.6+(3−h1)=10.3+18.6+3−h1P_1 = P_0 + 18.6 + (3 - h_1) = 10.3 + 18.6 + 3 - h_1 (metres of the column of water), we obtain from equation (1) a quadratic equation for h1h_1: h13=3×10.331.9−h1\frac{h_1}{3} = \frac{3 \times 10.3}{31.9 - h_1}, solving this, we get h1=1h_1 = 1 m approx. Thus the new (compressed) volume of air in the box equals approximately 1 m³. 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 m³, i.e., 1 tonne wt.

Browse the problem library