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A system of two blocks A and B and a wedge C are released from rest as shown in figure. Masses of the blocks and the wedge are $m,2m \;and \;2m$ respectively. The displacement of the wedge C when the block slides down the plane a distance 10 cm is (neglect friction)

\[\begin {array} {1 1} (a)\;5 \sqrt 2\;cm & \quad (b)\;4\;cm \\ (c)\;3 \sqrt 2\;cm & \quad  (d)\;\frac{5}{\sqrt 2}\;cm \end {array}\]

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X displacement of each block $=\large\frac{10}{\sqrt 2}$$ \qquad (10\;\cos 45)$
X displacement of wedge $=\large\frac{2m \times \Large\frac{10}{\sqrt 2}+m \times \Large\frac{10}{\sqrt 2}}{5 m}$
$\qquad= \large\frac{3}{5} \large\frac{10}{\sqrt 2}$
$\qquad= 3 \sqrt 2 \;cm$
answered Nov 22, 2013 by meena.p

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