Ground in the circuit of full wave rectifier
I just did this lab practice a few weeks ago, I think that you are asking about the circuit on this schematic:



A full wave rectifier based on center tap consists of two diodes in it as well as a center tapped transformer along with that a resistive load is connected across it.

The way it works is as follows:

The voltage provided by the AC Voltage source is driving the left inductor on the schematic, as it is magnetically coupled to the coil on the right the voltage across of the right coil is proportional to voltage across coil one time the transformer winding ratio, so if you have a ratio of lets say 2:1 the voltage across the second coil would be half the voltage across the first.


  Please tell me what is the reason behind it and why ground is not needed in bridge rectifier?


Ground is not needed because lets say the inductor on the right has a voltage drop of 10V. So you can see this in two different ways. the first is that one side of the coil is at 0V and the other at 10V, the other way is that one side is at 5V and the other at -5V. 

As the center tapping from the transformer is placed right in the middle we can tell that the voltage on the center tap would be half the voltage drop across the inductor, so if one side is at -5V and the other at 5V the center tap is at 0V potential. 

Keep in mind that this is a virtual ground relative to the coil voltage, because as you said it is not strictly at ground potential, but it works as ground for the load you are driving on the right side of the circuit.