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Ordinary is when you can physically stand on this said ground. Virtual ground occurs when the ground is virtually the same between two disagreeing parties. EX: I like cake and you like cake. That is virtual ground in an argument. Whereas, a disagreement would be an atheist going to Church. However that's a paradox...

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Q: What is ordinary ground and virtual ground in difference?
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What is the difference between ordinary ground And virtual ground?

Ordinary ground is a simple connection to ground. Virtual ground is an op-amp's response to maintaining a reference point in the bridge with respect to an ordinary ground. This is why an op-amp makes a nearly ideal mixer - it allows you to sum multiple inputs together with little or (nearly) no cross-talk. (Limitations, of course, being based on CMRR and frequency response.)


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Should op amp remain in active region for virtual ground?

An op amp must remain in its active region for virtual ground to be preserved.


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Which characteristics of Op-Amp is responsible for virtual ground?

The main parameter responsible for virtual ground of an Op-Amp is Input Impedance . Generally , for an Op-Amp it is very high . So , we can conclude that the characteristics of an Op-Amp responsible for virtual ground is high input impedance .


What is virtual ground and virtual short circuit?

VIRTUAL GROUND:-1]If the non-inverting (+) terminal of OP-AMP is connected to ground, then due to the "virtual short" existing between the two input terminals, the inverting (-) terminal also be at ground potential. hence it is said to be as "virtual ground".2]similarly if the inverting (-) terminal is connected to ground, then the non- inverting (+) terminal will be at "virtual ground" potential.3]The concept of virtual ground has been used extensively for amplifier analysis, specially we use this concept in the inverting amplifier.VIRTUAL SHORT:-1] The input impedance (Ri) of an OP-AMP is ideally infinite. Hence current "I" flowing from one input terminal to the other will be zero.2] Thus the voltage drop across Ri will be zero and both input terminals will be at the same potential. In other words they are virtually shorted to each other.