The raft, the anvil and the swimming pool.
Jill and a very large anvil are on a raft floating in the middle of a 10 m by 20 m rectangular swimming pool. The water is at a uniform height of 1.80 m. If Jill drops the anvil in the swimming pool, what will be the new height ( if it changes ) of the water in the swimming pool?
mass of anvil= 5000 kg
density of anvil= 5000 kg/m3
density of water= 1000 kg/m3
The answer is 1.78.
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1 solution
surface area of the swimming pool=20x10=200 m^2
volume of water in the swimming pool + the volume of the part of the raft under water=200x1.8=360 m^3
volume of water displaced by the anvil when on the raft V=Mass/density=5000/1000= 5 m^3
volume of water displaced by anvil when in the water V=mass/density=5000/5000 = 1 m^3
volume has decreased by (5-1)= 4 m^3
volume of water in the swimming pool +volume of raft under water+ anvil under water= 360-4= 356 m^3
new height = volume/surface=356/200= 1.78 m