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In the arrangement shown in the figure, the rod R is restricted to move in the vertical direction with acceleration $a_{1}$ and the block $B$ can slide down the fixed wedge with acceleration ${a_{2}}_{}$ . The correct relation between $a_{1}$ and $a_{2}$ is given by

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$a_{2} \cos θ = a_{1}$

$a_{2} \sin θ = a_{1}$

$a_{2} = a_{1} \sin θ$

$a_{2} = a_{1} \cos θ$

answer is B.

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Detailed Solution

$\frac{a_{1}}{a_{2}} = \sin θ$

$∴ a_{1} = a_{2} \sin θ$

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