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(D) Let the mass of each brick be $m$. The $x$-coordinate of the centre of mass of each individual brick is at its geometric centre.Taking the origin

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$\frac{22L}{35}$

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(D) Let the mass of each brick be $m$. The $x$-coordinate of the centre of mass of each individual brick is at its geometric centre.Taking the origin $O$ at the left edge of the bottom brick (brick $1$):$x_1 = \frac{L}{2}$$x_2 = \frac{L}{2} + \frac{L}{10}$$x_3 = \frac{L}{2} + \frac{2L}{10}$$x_4 = \frac{L}{2} + \frac{3L}{10}$$x_5 = \frac{L}{2} + \frac{2L}{10}$$x_6 = \frac{L}{2} + \frac{L}{10}$$x_7 = \frac{L}{2}$Since there are $7$ identical bricks,the total mass $M = 7m$.The $x$-coordinate of the centre of mass of the system is given by:$x_{cm} = \frac{\sum m_i x_i}{\sum m_i} = \frac{m(x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7)}{7m}$$x_{cm} = \frac{1}{7} [(\frac{L}{2}) + (\frac{L}{2} + \frac{L}{10}) + (\frac{L}{2} + \frac{2L}{10}) + (\frac{L}{2} + \frac{3L}{10}) + (\frac{L}{2} + \frac{2L}{10}) + (\frac{L}{2} + \frac{L}{10}) + (\frac{L}{2})]$$x_{cm} = \frac{1}{7} [7(\frac{L}{2}) + 2(\frac{L}{10} + \frac{2L}{10} + \frac{3L}{10})]$$x_{cm} = \frac{1}{7} [\frac{7L}{2} + 2(\frac{6L}{10})] = \frac{1}{7} [\frac{7L}{2} + \frac{6L}{5}]$$x_{cm} = \frac{1}{7} [\frac{35L + 12L}{10}] = \frac{47L}{70}$Wait,re-evaluating the displacement pattern from the image: The bricks are stacked symmetrically. The middle brick (brick $4$) is shifted by $3L/10$. The calculation above is correct based on the symmetry shown.

Seven identical homogeneous bricks,each of length $L$,are arranged as shown in the figure. Each brick is displaced with respect to the one in contact by $\frac{L}{10}$. Calculate the $x$-coordinate of the centre of mass of this system relative to the origin $O$ as shown.

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