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

Options

A. `4ams^(-2)`
B. `asqrt(17-8 cos alpha ) ms^(-2)`
C. `(sqrt(17))ams^(-2)`
D. `sqrt(17) cos.(alpha)/(2) xx a ms^(-2)`

Answer

B

Explanation

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 fig. that is
`|vec(a)_(BG)|=sqrt(q^(2)+(4a)^(2) + 2 xx a xx 4a xx cos (pi-alpha))`
`=(sqrt(17-8 cos alpha))a ms^(-2)`.