Q#33 (Past Exam Paper – March 2019 Paper 12 Q14)
An object shaped as a hemisphere rests with its flat surface on a table. The object has radius r and density ρ.
The volume of a sphere is 4/3 π r3.
Which average pressure does the object exert on the table?
A 1/3 ρr2 B 1/3 ρr2g C 2/3 ρr D 2/3 ρrg
Solution:
Answer: D.
The object is shaped as a hemisphere (half a sphere).
Pressure = Force / Area
Here, the force is the weight of the object and the area is the cross-sectional area of the object which is in contact with the table.
Volume of object = ½ × volume of sphere
Volume of object = ½ × 4/3 πr3 = 2/3 πr3
Density = mass / volume
Mass of object = ρV = 2/3 πρr3
Weight of object = mg = 2/3 πρr3g
The area in contact with the table has the shape of a circle of radius r.
Area = πr2
Pressure = Force (or weight) / Area
Pressure = (2/3 πρr3g) / (πr2)
Pressure = 2/3 ρrg
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