1 would also like to know what they mean by the maximum voltage by which the "virtual ground mode departs from its ideal value". The question is in the picture attached

 

They probably mean the deviation of the voltage from an ideal 0V at the negative input terminal of the Op-amp - assuming the positive input terminal is connected to supply ground.

At 0V input, the deviation would presumably be zero. The maximum deviation would occur when the input is either at +10V or -10V or both.

I guess one assumes that it is a unity gain inverting configuration.

 

They probably mean the deviation of the voltage from an ideal 0V at the negative input terminal of the Op-amp - assuming the positive input terminal is connected to supply ground.

At 0V input, the deviation would presumably be zero. The maximum deviation would occur when the input is either at +10V or -10V or both.

I guess one assumes that it is a unity gain inverting configuration.

I seriously doubt the reply given by our friend t n k. There is no need to assume a unity gain or not. We know that Vout=A(open loop gain)*vd(voltage across 2nd and 3rd pin) irrespective of whether the opamp is in open loop or closed loop. So here the virtual ground node is nothin but the 2nd pin. The 3rd pin is connected stiffly to gnd. So in order to produce +10 V as output, the vd=Vout/A. Thus vd=10/2000=5mV. Here vd=Vin+ - Vin- = 0 - Vin-. So voltage at virtual gnd must be -5mV for this case. For Vout=-10 V it is vice versa ie voltage at 2nd pin is +5mV. I hope I make sense

 

I 'll agree with elecidiod on that one.

 

I agree as well.

I was mistakenly thinking the input to an inverting stage was +10V to -10V rather than the output being +10V to -10V.