A square hole is punched out of a circular lamina, the diagonal of the square being the radius of the circle. If ‘a’ be the diameter of the circle, find the distance of C.M. of the remaining portion from the centre of the circle.

The problem involves the concept of center of mass in physics and geometry. Here are the steps to solve it:

Step 1: Understand the problem A square hole is punched out of a circular lamina. The diagonal of the square is the radius of the circle. We need to find the distance of the center of mass (C.M.) of the remaining portion from the center of the circle.

Step 2: Define the variables Let 'a' be the diameter of the circle. Therefore, the radius 'r' of the circle is a/2. Since the diagonal of the square is equal to the radius of the circle, the side 's' of the square is r/√2 = a/(2√2).

Step 3: Calculate the areas The area of the circle 'Ac' is πr² = π(a/2)² = πa²/4. The area of the square 'As' is s² = (a/(2√2))² = a²/8.

Step 4: Calculate the area of the remaining portion The area of the remaining portion 'Ar' is Ac - As = πa²/4 - a²/8 = (π - 1/2)a²/4.

Step 5: Calculate the distance of the C.M. The distance of the C.M. of the remaining portion from the center of the circle is given by the formula d = (Acr - Asd')/Ar, where d' is the distance of the C.M. of the square from the center of the circle. Since the square is symmetric, d' = 0. Therefore, d = (Ac*r)/Ar = (πa²/4 * a/2) / ((π - 1/2)a²/4) = πa/(2(π - 1/2)).

So, the distance of the C.M. of the remaining portion from the center of the circle is πa/(2(π - 1/2)).