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(D) Let the mass of the horizontal rod be $m_1$ and the mass of the vertical rod be $m_2$.The centre of mass of the horizontal rod is at $(L/2, 0)$ an

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$(L/3, L/6)$

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(D) Let the mass of the horizontal rod be $m_1$ and the mass of the vertical rod be $m_2$.The centre of mass of the horizontal rod is at $(L/2, 0)$ and the centre of mass of the vertical rod is at $(0, L/2)$.The coordinates of the centre of mass $(X_{cm}, Y_{cm})$ of the system are given by:$X_{cm} = \frac{m_1(L/2) + m_2(0)}{m_1 + m_2} = \frac{m_1 L}{2(m_1 + m_2)}$$Y_{cm} = \frac{m_1(0) + m_2(L/2)}{m_1 + m_2} = \frac{m_2 L}{2(m_1 + m_2)}$Since the rods are made of different materials,$m_1 
eq m_2$. Let $\lambda = m_2/m_1$. Then $X_{cm} = \frac{L}{2(1+\lambda)}$ and $Y_{cm} = \frac{\lambda L}{2(1+\lambda)}$.Note that $X_{cm} + Y_{cm} = \frac{L}{2(1+\lambda)} + \frac{\lambda L}{2(1+\lambda)} = \frac{L}{2}$.Checking the options,only $(L/3, L/6)$ satisfies $X_{cm} + Y_{cm} = L/2$ (i.e.,$L/3 + L/6 = L/2$).Thus,the correct option is $(L/3, L/6)$.

The centre of mass of two thin uniform rods of the same length $L$ but made of different materials and kept as shown in the figure can be,if the meeting point is the origin of coordinates:

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