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(C) Let the radii of the three discs be $r_1 = r$, $r_2 = 2r$, and $r_3 = 3r$. Since they are made of the same material and have the same thickness, t

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$6$

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(C) Let the radii of the three discs be $r_1 = r$, $r_2 = 2r$, and $r_3 = 3r$. Since they are made of the same material and have the same thickness, their masses are proportional to their areas: $m \propto \pi r^2$.Thus, $m_1 = k(\pi r^2) = m$, $m_2 = k(\pi (2r)^2) = 4m$, and $m_3 = k(\pi (3r)^2) = 9m$.Let the centre of the smaller disc $(m_1)$ be at the origin $(0, 0)$.The middle disc $(m_2)$ touches the first disc, so its centre is at $x_2 = r + 2r = 3r$.The third disc $(m_3)$ touches the middle disc, so its centre is at $x_3 = x_2 + 2r + 3r = 3r + 5r = 8r$.The centre of mass $X_{cm}$ is given by:$X_{cm} = \frac{m_1x_1 + m_2x_2 + m_3x_3}{m_1 + m_2 + m_3}$$X_{cm} = \frac{m(0) + 4m(3r) + 9m(8r)}{m + 4m + 9m} = \frac{12mr + 72mr}{14m} = \frac{84mr}{14m} = 6r$.The distance of the centre of mass from the centre of the smaller disc is $6r$.

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