__Question 33__
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 r

^{3}.
Which average pressure does the object exert on the
table?

**A**1/3 ρr

^{2}

**B**1/3 ρr

^{2}g

**C**2/3 ρr

**D**2/3 ρrg

**Reference:**

*Past Exam Paper – March 2019 Paper 12 Q14*

**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 πr

^{3}= 2/3 πr^{3}
Density = mass / volume

Mass of object = ρV = 2/3 πρr

^{3}
Weight of object = mg = 2/3 πρr

^{3}g
The area in contact with
the table has the shape of a circle of radius r.

Area = πr

^{2}
Pressure = Force (or weight)
/ Area

Pressure = (2/3 πρr

^{3}g) / (πr^{2})
Pressure = 2/3 ρrg