A part was lowered into a cylindrical vessel containing 6 dm3 of water. At the same time, the level of the liquid in the vessel increased 1.5 times. What is the volume of the part?
Let the plane of the base of the cylinder be Sbn dm2, and the height of the liquid in the vessel is h cm.
Then the volume of liquid in the vessel will be equal to:
V1 = Sbn * h cm3 = 6 dm3.
After immersion in the vessel of the part, the liquid level became equal to 1.5 * h dm.
Since the liquid level has risen 1.5 times, the volume will also increase 1.5 times.
V2 = S main * 1.5 * h = 6 * 1.5 = 9 dm3.
Then the difference in volumes is the volume of the immersed part.
V = V2 – V1 = 9 – 6 = 3 dm3.
Answer: The volume of the part is 3 dm3.
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