A cylindrical disk of wood weighing 48.0 N and having a diameter of 30 cm floats

Question

A cylindrical disk of wood weighing 48.0 N and having a diameter of 30 cm floats on a cylinder of oil of density 0.850 g/cm^3 (the figure ). The cylinder of oil is 75.0 cm deep and has a diameter the same as that of the wood.
What is the gauge pressure at the top of the oil column? 679 Pa
Suppose now that someone puts a weight of 95.0 on top of the wood, but no oil seeps around the edge of the wood. What is the change in pressure at the bottom of the oil? 1340 Pa

What is the change in pressure at the halfway down in the oil?
= ? Pa

Explanation / Answer

Guage pressure at top = Weight/area = 48.0N/{Pi*0.15m^2} = 679Pa Change in pressure at the bottom = added weight/area = 95.0N/{Pi*0.15m^2} = 1344Pa Change in pressure at the middle, sort of a trick question, the change in pressure is just the added weight divided by the area. This change in pressure is the same everywhere in the column, so still 1340 Pa

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