• # question_answer 42) In an arrangement shown in the figure, the acceleration of block A and B are given- A) g/3, g/6             B) g/6, g/3C) g/2, g/2             D) 0, 0

$\,T-mg\text{ }sin\,30{}^\circ =m\left( 2a \right)$ $or\,\,\,T-\frac{mg}{2}=2ma\,\,\,\,\,\,\,\,\,\,\,\,......(1)$ $and\text{ }mg-2T=ma\text{ }\,\,\,\,\,\,\,\,\,\,....\left( 2 \right)$ Solving equation (1) and (2) we get $a=0$