@AnkitKumar As DavidG25 indicates, if an ideal voltage source is applied directly to the two inputs of the opamp, then the output of the opamp will be its open loop voltage gain times that ideal voltage source's value (assuming the opamp is otherwise ideal -- no input bias or offset, etc.) If you got the schematic accurately copied out, then you probably don't understand what's connected at the two input leads and need to uncover the intent of this circuit.
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