• # question_answer If the system is released, then the acceleration of the centre of mass of the system (as shown in the figure) is: A)  $\frac{g}{4}$                                   B) $\frac{g}{2}$C)  $g$                            D)  $2g$

Let the magnitude of acceleration of the block be a. $a=\frac{3mg-mg}{3m+m}=\frac{g}{2}$ Let the downward direction be positive and upward direction negative. Now, ${{a}_{cm}}=\frac{{{m}_{1}}{{a}_{1}}+{{m}_{2}}{{a}_{2}}}{{{m}_{1}}+{{m}_{2}}}=\frac{3m\frac{g}{2}-\frac{mg}{2}}{3m+m}$ $\therefore$    ${{a}_{cm}}=\frac{mg}{4m}=\frac{g}{4}$ Hence, the correction option is [a].