Let the minimum edge of the block = X cm
The volume of the ice block =X³ cm³ =X³/10⁶ m³
Density of the ice = 0.9*1000 kg/m³ =900 kg/m³
The weight of the ice block =(X³/10⁶)*900 kg
Total weight of the ice block and the metal piece = (X³/10⁶)*900+0.50 kg
The force of the buoyancy = weight of the water displaced =(X³/10⁶)*1000 kg. (The volume of water displaced just before it sinks is equal to the volume of the ice cube = X³ cm³)
In the given condition the weight of ice cube and the metal = force of buoyancy,
→(X³/10⁶)*900+0.50 = (X³/10⁶)*1000
→(X³/10⁶)*(1000-900) =0.50
→X³ = 0.50*10⁶/100 =5000
→X ≈ 17 cm
