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(C) Let the linear mass density of the rods be $\lambda$.The mass of the horizontal rod (length $L$) is $m_1 = \lambda L$ and its centre of mass is at

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$\left(\frac{L}{6}, \frac{2 L}{3}\right)$

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(C) Let the linear mass density of the rods be $\lambda$.The mass of the horizontal rod (length $L$) is $m_1 = \lambda L$ and its centre of mass is at $(x_1, y_1) = (L/2, 0)$.The mass of the vertical rod (length $2L$) is $m_2 = \lambda(2L) = 2m$ (where $m = \lambda L$) and its centre of mass is at $(x_2, y_2) = (0, L)$.The $x$-coordinate of the centre of mass is $x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} = \frac{m(L/2) + 2m(0)}{m + 2m} = \frac{mL/2}{3m} = \frac{L}{6}$.The $y$-coordinate of the centre of mass is $y_{cm} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2} = \frac{m(0) + 2m(L)}{m + 2m} = \frac{2mL}{3m} = \frac{2L}{3}$.Thus,the coordinates of the centre of mass are $\left(\frac{L}{6}, \frac{2L}{3}\right)$.

The figure shows a composite system of two uniform rods of lengths as indicated. The coordinates of the centre of mass of the system of rods are ...........

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