A cylindrical disk of wood weighing 50.0 Ans having a diameter of 30 cm floats o

Question

A cylindrical disk of wood weighing 50.0 Ans having a diameter of 30 cm floats on a cylinder of oil of density of 0.85 g/cm3. The Cylander of oil is 75 cm deep and has a diameter the same as that of the wood.
Part A) What is the pressure at the top of the column? (In Pa)
Part B) Supposed now that someone puts a weight of 95 N on top of the wood, but no oil seeps around the edge of the wood. What is the CHANGE in pressure? (In Pa)
Part C) What is the CHANGE in pressure at halfway down the oil? (In Pa)

Explanation / Answer

P=F/A=50N/(pi*(0.015^2)=70735.53 Pa

because A=pi*r^2 and radius=(1/2)diameter

b) (95N+50N)/(pi*(0.015^2)=205133.0378Pa

c) pressure = 205133.0378Pa + h*rho*g =205133.0378Pa + 0.75m/2 *850*9.81

pressure =208259.9753 N

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