A hanging mass $$M$$ is connected to a four times bigger mass by using a string pulley arrangement, as shown in the figure. The bigger mass is placed on a horizontal ice-slab and being pulled by $$2Mg$$ force. In this situation, tension in the string is $$\frac{x}{5}Mg$$ for $$x =$$ ______. Neglect mass of the string and friction of the block (bigger mass) with ice slab.

Let us start by considering the forces on the 4M block

There is an external Force pulling it which is $$F=2Mg$$ and $$T\left(tension\ of\ the\ rope\right)$$

we know that here masses M and 4M have the same acceleration , lets take it as 'a'

now writing equation for the 4M block

$$2Mg\ -\ T\ =4Ma$$

and for the M block 

Forces at $$T\left(tension\ of\ the\ rope\right)$$ and $$F=Mg$$(due to gravity)

now writing equation for M block

$$T-Mg=Ma$$

now on adding these both equations

$$2Mg-Mg=5Ma$$

$$Mg=5Ma$$

$$\frac{g}{5}=a$$

Now substituting value of a in equation 1

$$2Mg\ -\ T\ =4M\times\ \frac{g}{5}$$

$$2Mg\ -\frac{4Mg}{5}\ =T$$

$$\frac{\left(10Mg\ -4Mg\right)}{5}\ =T$$

$$\frac{6Mg}{5}\ =T$$

Therefore x=6