Why is the output DC voltage of full wave rectifier not equal to the RMS of the original AC input?
The average of a full-wave rectified sine wave of peak voltage A is 2A/\$\pi\$

The RMS value of a full-wave rectified sine wave of peak voltage A is A/\$\sqrt{2}\$

You can easily verify these numbers by computing the definite integral of the relevant function over a half cycle. 

So the ratio of average to RMS (for a sine wave!) is 2\$\sqrt{2}/\pi\$ or about 0.9003163

Consider a 10VDC supply going into a 10 ohm resistor. Average = RMS = 10VDC and power is 10W. 

Now consider a 1000V pulse with 1% duty cycle. The average is 10V, however the RMS is 100V and the power is 1000W (100kW for 1% of the time). 

The "spikier" the waveform the higher RMS is to average.