1. Rule to decide gas evolved at anode

At the anode, oxidation occurs. We check which species is more easily oxidized:

• the anion/water in solution, or
• the electrode itself, if it is an active electrode like Ag.

If water is oxidized at inert electrodes, oxygen is formed: $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$ (or equivalently in basic form).

2. Case (A): Aqueous $\mathrm{AgNO_3}$ using silver electrodes

Here the electrode is silver, which is an active electrode. At the anode, silver itself gets oxidized more readily: $Ag \rightarrow Ag^+ + e^-$ So the anode dissolves, and $O_2$ is not evolved.

Thus, (A) is false.

3. Case (B): Aqueous $\mathrm{AgNO_3}$ using platinum electrodes

Platinum is inert, so the electrode does not dissolve. Possible oxidation at anode:

• $NO_3^-$ is very difficult to oxidize further.
• Water gets oxidized instead.

Hence oxygen evolves: $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$

Thus, (B) is true.

4. Case (C): Dilute $\mathrm{H_2SO_4}$ using platinum electrodes

With inert Pt electrodes in dilute sulfuric acid:

• At cathode: $H^+$ is reduced to $H_2$.
• At anode: water is oxidized to $O_2$.

Anode reaction: $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$

Thus, (C) is true.

5. Case (D): High concentration solution of $\mathrm{H_2SO_4}$ using platinum electrodes

In highly concentrated sulfuric acid, sulfate/bisulfate participates and the usual electrolysis does not simply give oxygen as the anode product in the standard JEE treatment. Oxygen evolution is characteristic for dilute sulfuric acid, not concentrated sulfuric acid.

Thus, (D) is false.

6. Conclusion

The cases in which $O_2$ is evolved are: $\boxed{(B)\text{ and }(C)\text{ only}}$

So the correct option is: $\boxed{A}$