A system is shown in the figure, which is released from rest. There is no friction between ground and mass 5m. The coefficient of friction between mass 5m and mass 2m is $\mu$ . The coefficient of friction between mass 5m and mass m is also  $\mu$. Choose the correct statement(s)

Since system has no external force acting along horizontal direction, so component of acceleration of centre of mass along horizontal direction is zero. Let magnitude of acceleration of mass m and mass 2m with respect to mass 5m be a.
Let acceleration of mass 5m is  ${\overrightarrow{A}}_{5m}=A\left(\widehat{i}\right)$
${\overrightarrow{A}}_m=a(-\widehat{i})+A\left(\widehat{i}\right)$ and ${\overrightarrow{A}}_{2m}=\frac{1}{2}a\left(\widehat{i}\right)+\frac{\sqrt{3}}{2}a(-\widehat{j})+A\left(\widehat{i}\right)$ , hence $\overrightarrow{0}=\frac{5mA\left(\widehat{i}\right)+2m[\frac{1}{2}a\left(\widehat{i}\right)+A\left(\widehat{i}\right)]+m[a(-\widehat{i})+A\left(\widehat{i}\right)]}{8m}\Rightarrow A=0$
 
Using FBD of mass 2m, we have
$2mg\sin {60}^0-\mu 2mg\cos {60}^0-T=2ma$ ………………(1)
Using FBD of mass m, we have
 $T-\mu mg=ma$………………(2)
Adding equations (1) and (2), we have
 
 $mg(2\sin {60}^{\circ }-2\mu \cos {60}^{\circ }-\mu )=3ma\Rightarrow a=\frac{g}{3}(\sqrt{3}-2\mu )=\frac{g}{3\sqrt{3}}$$T=ma+\mu mg=\frac{mg}{3\sqrt{3}}+\frac{mg}{\sqrt{3}}=\frac{4mg}{3\sqrt{3}}$