Let the mass of the lead piece =X g
Total weight of the wood and the lead = 200+X g
The volume of the lead piece = X/11.3 cm³
{Since the density of lead = 11.3 g/cm³}
When the wooden block is just allowed to float in water, it displaces a volume of water equal to its volume = 200/0.8 cm³ =250 cm³
{Because the density of wood =0.8*1 g/cm³ =0.8 cm³}
In the given condition total volume of the water displaced =(250 + X/11.3) cm³
Hence the force of buoyancy
=(250 + X/11.3) cm³ * 1 g/cm³
=(250 + X/11.3) g
{NOTE: In these problems, the unit of force is taken as gram-weight or kg-weight, not dyne or Newton because while equating, the conversion factor g, the acceleration due to gravity, cancel out. So do not be confused}
Now while just allowed to float on the water, the weight of wood plus the lead piece = the force of buoyancy.
→ 200+X = (250 + X/11.3)
→ X-X/11.3 =250-200
→10.3X =50*11.3
→X = 50*11.3/10.3
= 54.8 g
