Solution:

Given, diameter (D) of the circular plate = 28 cm and diameter (d) of the removed portion = 21 cm

Let the thickness of the plate be t.

Let the distance of the centre of mass of the remaining portion from the centre of the plate be x.

Step 1: Finding the Centre of Mass of the Plate

The centre of mass of a circular plate lies at its centre.

Hence, the centre of mass of the given plate lies at point O.

Step 2: Finding the Mass of the Plate

The mass of the plate can be calculated using its volume and density.

The volume of the plate can be calculated as follows:

Volume of plate = π/4 × D² × t

= π/4 × 28² × t

= 196πt cm³

Given that the plate has uniform thickness, its density will also be uniform.

Let the density of the plate be ρ.

Then, the mass of the plate can be calculated as follows:

Mass of plate = Density × Volume

= ρ × 196πt grams

Step 3: Finding the Centre of Mass of the Remaining Portion

Since the circular portion of diameter 21 cm is removed from the plate, the remaining portion can be considered as a ring of thickness t and radius R = 14 cm.

The centre of mass of a ring lies on its axis of symmetry.

Hence, the centre of mass of the remaining portion lies on the line passing through the centres of the plate and the removed portion.

Let the distance of the centre of mass of the remaining portion from the centre of the plate be x.

Then, the distance of the centre of mass of the removed portion from the centre of the plate will be (14 - x).

The mass of the remaining portion can be calculated as follows:

Mass of remaining portion = Density × Volume

= ρ × π/4 × [R² - (d/2)²] × t grams

= ρ × π/4 × [14² - 10.5²] × t grams

= 33.25πρt grams

The centre of mass of the remaining portion can be calculated using the following formula:

x = (m1x1 + m2x2) / (m1 + m2)

where m1 and x1 are the mass and position of the centre of mass of the plate, and m2 and x2 are the mass and position of the centre of mass of the remaining portion.

Substituting the values, we get:

x = [ρ × 196πt × 0 + 33.25πρt × (14 - x)] / [ρ × 196πt + 33.25πρt]

Simplifying the above equation, we get:

x = 33.25 / 77.25 × 14

= 5.99 cm

Hence, the position of the centre of mass of the remaining portion will shift towards left from ‘O’ by 5.99 cm.