A square shaped hole of side l = a/2 is carved out at a distance d = a/2 from the centre ‘O’ of a uniform circular disk of radius a. If the distance of the centre of mass of the remaining portion form O is - a/X , value of X (to the nearest integer) is :

Solution

Physics sir · 4 steps

Treat the hole as negative mass: full disk minus the square, both measured from O.

1. Negative mass idea. The remaining plate is the full disk with the square taken away. Give the square a negative mass. Then use the usual centre of mass formula along the line through O and the square's centre.
2. Masses and positions. Take mass proportional to area. The full disk has area π a² and its centre is at x = 0. The square has area (a/2)² = a²/4 and its centre is at x = a/2.

   x_(cm) = (π a² × 0 - a²/4×a/2)/(π a² - a²/4) = -(a/8)/(π - 1/4) = -a/(8π - 2)

3. Put in numbers. 8π = 25.13, so 8π - 2 = 23.13. The minus sign means the centre of mass moves away from the hole, to the opposite side of O.
4. So. x_(cm) = -a/23.13, so X rounds to 23.

The common slip: 25. Leaving the denominator as the full disk area π a² instead of π a² - a²/4 gives X = 8π ≈ 25, not 23.

The idea to keep. For a plate with a hole cut out, treat the hole as a negative mass. Subtract its moment on top and its mass below. The centre of mass always moves away from the hole.

Answer from the NTA final answer key. Solution by PrepWiser.