In the diagram shown, the acceleration of the block `B` as shown in relative to the block A and relative to ground is `a_(BA)` and `a_(BG)` respectively. If the block A is moving towards left with an acceleration `a_(0)` then
.

Options

A. `a_(BA) =2a_(0)`
B. `a_(BG) =3a_(0)`
C. `a_(BA) =3a_(0)`
D. `a_(BG) = a_(0) sqrt(10 + 6 cos theta)`

Answer

C, D

Explanation

Let `T` be the tension, `3Ta_(0)-Ta_(BA) =0`
`a_(BA)=3a_(0),a_(BG)=sqrt(a_(0)^(2)+(3a_(0))^(2)+2.a_(0).3a_(0)"cos"theta)`
`a_(BG)=a_(0)sqrt(10+6c"co"stheta)` .