A rod of length 5 L is bent right angle keeping one side length as 2 L .
The position of the centre of mass of the system :
(Consider L=10 cm)

Solution

Physics sir · 4 steps

Treat each arm as a point mass at its midpoint, weighted by its length. The bend changes nothing else.

1. Split the rod. The rod is uniform, so each arm's mass is proportional to its length. The vertical arm has length 3L and the horizontal arm has length 2L, so take their masses as 3m and 2m.
2. Locate each midpoint. The vertical arm lies along the y-axis from 0 to 3L, so its centre is at (0, 1.5L). The horizontal arm lies along the x-axis from 0 to 2L, so its centre is at (L, 0).

   x_(cm) = (3m(0) + 2m(L))/5m = 2L/5, y_(cm) = (3m(1.5L) + 2m(0))/5m = 9L/10

3. Put in L. With L = 10 cm, x_(cm) = (2 × 10)/5 = 4 cm and y_(cm) = (9 × 10)/10 = 9 cm.
4. So. The centre of mass is at 4î + 9ĵ (in cm), which is option (a).

The idea to keep. For a bent uniform rod, replace each straight piece by its mass at its midpoint, with mass in proportion to length, then take the weighted average of the coordinates.

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