A rod of length 5L is bent right angle keeping one side length as 2L.

The position of the centre of mass of the system: (Consider L = 10 cm)

Let the rod be uniform, so its linear mass density is the same everywhere. Total length = $$5L$$. After bending it at a right angle we get two straight sections that are mutually perpendicular:

• One section has length $$2L$$. • The other section has length $$3L$$ (because $$2L + 3L = 5L$$).

Choose a Cartesian frame with the bend (the joint) at the origin. Place the $$2L$$ section along the +x-axis and the $$3L$$ section along the +y-axis.

Let $$\lambda$$ be the linear mass density. Mass of the x-section: $$m_1 = \lambda \,(2L)$$. Mass of the y-section: $$m_2 = \lambda \,(3L)$$. Total mass: $$M = m_1 + m_2 = \lambda\,(5L)$$.

Centre of mass along x-axis
Only the x-section contributes. Its centre lies halfway along its length, i.e. at $$x = \frac{2L}{2} = L$$. Hence

$$x_{cm} = \frac{m_1 \,(L)}{M} = \frac{\lambda\,(2L)\,L}{\lambda\,(5L)} = \frac{2L}{5}$$

Centre of mass along y-axis
Only the y-section contributes. Its centre lies at $$y = \frac{3L}{2} = 1.5L$$. Hence

$$y_{cm} = \frac{m_2 \,(1.5L)}{M} = \frac{\lambda\,(3L)\,(1.5L)}{\lambda\,(5L)} = \frac{4.5L}{5} = \frac{9L}{10}$$

Now substitute $$L = 10 \text{ cm}$$:

$$x_{cm} = \frac{2}{5}\,(10\,\text{cm}) = 4 \,\text{cm}$$
$$y_{cm} = \frac{9}{10}\,(10\,\text{cm}) = 9 \,\text{cm}$$

Therefore the position vector of the centre of mass is $$\mathbf{r}_{cm} = 4\hat{i} + 9\hat{j}\,\text{cm}$$.

Option D is correct.