Does the virtual ground condition hold true when I cut off the power supply of the opamp?
I am using an opamp as a buffer. If I cut off the power supply but continue to give a signal on the non-inverting (+) input, will virtual ground work and will I get the output?
Does the virtual grd condition hold true when I cut off the power supply of the opamp?
No, a virtual ground relies on the op-amp having power.
if I cut off the power supply but continue to give a signal on the + signal pin will virtual grd work and will I get the output?
No, and maybe you'll get something on the output but it won't be what you want and you might destroy the op-amp by applying an input when the power is removed.
"Virtual ground" means that the opamp is driving the signal to ground level. But you only get a "virtual ground" in inverting configurations whereas a "buffer" is non-inverting. So it is not even clear what kind of circuit you are talking about.
When a virtual ground is part of the circuit, the opamp actively drives the signal to it. Obviously that requires power rails to work with.
Many opamps have different kind of input protection circuits like clamping diodes to the power rails and/or between the inputs. Those are likely to engage when the opamp is powered off and carry some current. Your spec sheet will tell you the maximum permissible amounts here.
As the power rails come down, the input will likely get outside of the voltage interval between the power rails that constitute valid input. That may lead to phenomena like phase inversion, with the output going hard to the wrong direction, pretty much the opposite of a "virtual ground".
This can cause significant noises when amps are being switched off.
In a broad sense of the word, virtual ground is a ubiquitous phenomenon that can be observed all around us. Its essence is the compensation of losses (minus) by adding an equivalent gain (plus); the result is a constant value (in particular, zero). Unlike ordinary zero, this "virtual zero" is a difference between two large quantities. For example, if you have 1000$ (your "virtual money zero") and spend 100$, you should earn 100$ to compensate for the loss (1000$ - 100$ + 100$ = 1000$). If you do not earn and only spend money, your amount ceases to be 1000$ and will constantly decrease (the OP's problem).
A voltage divider R1-R2 supplied on both sides with opposite polarity voltages Vin and -Vout is a typical virtual ground arrangement. The input voltage source Vin creates a voltage drop Vin.R2/(R1 + R2) at the middle point. As a reaction, the negative output voltage source Vout creates a "counter voltage drop" -Vin.R1/(R1 + R2) at the middle point to make it zero. The two voltages are summed (superposition) and when V2/V1 = -R2/R1, the result is zero (virtual ground).
If Vout disappears (0 V), the virtual ground ceases to be zero and begins to follow Vin according to Vin.R2/(R1 + R2); this is the OP's problem. Here we assume that Vout has zero internal resistance.
Now all that remains is to make an op-amp do the work of Vout and we get the circuit of the classic op-amp inverting amplifier. It is clear that if we turn off the power supply (the OP's question), the "magic" of the virtual ground disappears and it starts to follow Vin according to Vin.R2/(R1 + R2). Here we assume that the op-amp has zero internal resistance when the power supply is disconnected.
