Seven homogeneous blocks each of length L are arranged as shown in the figure. Each block is displaced with respect to the one contact by L/10.The X-coordinate of the center of mass of the system relative to the origin O is

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$\frac{22 L}{35}$

$\frac{11 L}{35}$

$\frac{11 L}{12}$

$\frac{22 L}{25}$

answer is C.

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Detailed Solution

$m_{1} = m (\frac{L}{2}, 0)$

$m_{2} = m (\frac{6L}{10}, 0)$

$m_{3} = m (\frac{7L}{10}, 0)$

$m_{4} = m (\frac{8L}{10}, 0)$

$m_{5} = m (\frac{7L}{10}, 0)$

$m_{6} = m (\frac{6L}{10}, 0)$

$m_{7} = m (\frac{L}{2}, 0)$

$y_{cm} = \frac{m (\frac{L}{2} + \frac{6L}{10} + \frac{7L}{10} + \frac{8L}{10} + \frac{7L}{10} + \frac{6L}{10} + \frac{L}{2})}{7 m}$

$y_{cm} = \frac{(L + \frac{34}{16} L)}{7} = \frac{44 L}{70} = \frac{22 L}{35}$

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