A cubical block of wood weighing 200 g has a lead piece fastened underneath. Find the mass of the lead piece which

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