question_answer

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/3

C) g/2, g/2             

D) 0, 0

Correct Answer: D

Solution :

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