In the arrangement shown, the strings are mass less, and the radii of the pulleys are: ${\text{r}}_{\text{1}}{\text{=3cm,r}}_{\text{2}}{\text{=6cm,r}}_{\text{3}}{\text{=4cm\hspace{0.17em}and\hspace{0.17em}r}}_{\text{4}}\text{=10cm}$ . Friction is absent everywhere and the angle ${\text{θ=sin}}^{\text{-1}}\left(\frac{\text{4}}{\text{5}}\right)$. The value of $\frac{\text{M}}{\text{m}}$ such that the system remains in equilibrium is ____________.

Starting from the string directly attached to $M,$ let the tension in the string be $T_1,T_2$ and  $T_3$
Balancing torque on the pulley’s,
$T_1{r}_4=T_2{r}_3\Rightarrow Mg{r}_4=T_2{r}_{3 }\mathrm{...}\left(i\right)$   
 $T_2{r}_2=T_3{r}_1 \Rightarrow T_2{r}_2=\left(mg\sin \theta \right)r_1 \mathrm{...}\left(ii\right)$  
Multiplying the two equations, we get
$\left(Mg\right)\left(r_4{r}_2\right)=\left(mg\sin \theta \right)\left(r_3{r}_1\right)\Rightarrow \frac{M}{m}=\frac{r_3{r}_1}{r_4{r}_2}\sin \theta =\frac{\left(4\right)\left(3\right)}{\left(10\right)\left(6\right)\left(5\right)}=0.16$