A 0.5 kg mass of lead is submerged in a container filled to the brim with water and a block of wood floats on top. The lead mass is slowly lifted from the container by a thin wire and as it emerges into air the level of the water in the container drops a bit. The lead mass is now placed on the block of wood. As the lead is placed on the wood.
Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.
Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA
When the lead is submerged, it displaces a volume of water equal to its own volume (V_lead). When lifted into air, it displaces no water. When placed on the floating wood, the combined system displaces a volume of water equal to the weight of both objects (Archimedes' principle). The weight of water displaced equals the total weight: m_wood*g + m_lead*g.
The volume displaced is V_displaced = (m_wood + m_lead)/ρ_water.
Since lead is denser than water (ρ_lead > ρ_water), its weight is greater than the weight of water equal to its volume (m_lead*g > ρ_water*V_lead*g). Therefore, V_displaced > V_lead + V_displaced_wood_alone. This means more water is displaced than when the lead was submerged and the wood floated separately. Consequently, the water level rises higher than before, and some water must spill over.
Final Answer: Some water spills over the edge of the container.