A cubical block of wood weighing 200 g has a lead piece fastened underneath. Determine the mass of the lead piece required to just allow the block to float in water. The specific gravity of wood is 0.8 and that of lead is 11.3 .

Step-by-step solution

Let the mass of the lead piece be m . Volume of the wooden block, V_w = 200 0.8 = 250 cm ^3 Volume of the lead piece, V_l = m 11.3 cm ^3 For the block to just float in water, the entire system (wood and lead) must be completely submerged. By the principle of flotation, the total mass of the system equals the mass of the displaced water. 200 + m = (V_w + V_l ) _ water Substituting the values, with _ water = 1 g/cm ^3 : 200 + m = (250 + m 11.3 ) 1 m - m 11.3 = 250 - 200 m ( 11.3 - 1 11.3 ) = 50 m ( 10.3 11.3 ) = 50 m