If the acceleration of wedge in the shown arrangement is a ms^-2towards left, then at this instant, acceleration of the block (magnitude only) would be

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$4 a {ms}^{- 2}$

$a \sqrt{17 - 8 \cos α} {ms}^{- 2}$

$(\sqrt{17}) a {ms}^{- 2}$

$\sqrt{17} \cos \frac{α}{2} \times a {ms}^{- 2}$

answer is B.

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Detailed Solution

If the wedge moves leftward by x, then the block moves down the wedge by 4x, i.e., w.r.t. wedge the block comes down by 4x. So, acceleration of block w.r.t wedge = 4a along the incline plane of wedge. Acceleration of wedge with respect to ground= a, along left. So acceleration of block w.r.t. ground is the vector sum of the two vectors shown in the figure. That is

$\begin{array}{l} |{\vec{a}}_{B G}| = \sqrt{a^{2} + (4 a)^{2} + 2 \times a \times 4 a \times \cos (π - α)} \\ = (\sqrt{17 - 8 \cos α}) a {ms}^{- 2} \end{array}$