A 0.5 kg mass of lead is submerged in a conta | vedclass

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(B) Initially,the lead mass is submerged,so it displaces a volume of water equal to its own volume $(V_{lead})$. The wood block floats,displacing wate

Vedclass · Accepted Answer

The water level rises exactly to the brim as before.

Vedclass · Answer

(B) Initially,the lead mass is submerged,so it displaces a volume of water equal to its own volume $(V_{lead})$. The wood block floats,displacing water equal to its weight $(W_{wood}/ho_{water}g)$.When the lead is removed,the water level drops because the volume $V_{lead}$ is no longer displaced.When the lead is placed on the wood,the total weight supported by the water increases by the weight of the lead $(W_{lead} = m_{lead}g)$.According to Archimedes' principle,the wood block must displace an additional volume of water equal to the weight of the lead divided by the density of water: $\Delta V = W_{lead} / (ho_{water}g) = m_{lead} / ho_{water}$.Since the density of lead $(ho_{lead} \approx 11.3 	imes 10^3 \ kg/m^3)$ is much greater than the density of water $(ho_{water} = 1.0 	imes 10^3 \ kg/m^3)$,the volume of water displaced by the weight of the lead $(m_{lead}/ho_{water})$ is significantly larger than the volume of the lead itself $(m_{lead}/ho_{lead})$.Therefore,the water level rises higher than it was when the lead was submerged,but since the container was initially full and water spilled out when the wood was first placed,the level will rise back to the brim.

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