Voltage gain and output voltage are related by the formula $A _v=\dfrac{output \ voltage}{input \ voltage}$. Substituting the value of input voltage, we get, output voltage = - 20 x 20mV = -400 mV

 An ideal op-amp has infinite bandwidth. Therefore, any frequency signal from 0 to $\infty$ Hz can be amplified without attenuation.

The ideal opamp voltage is maintained constant. It is controlled by the input voltage. Thus the output of an opamp is controlled by the source (input) voltage

In an ideal opamp, output follows the input instantly and hence the time delay is zero. slew rate is calculated as the ratio of change in voltage at the output to time taken for this action to take place. If time delay is zero, the slew rate will be infinite

maximum voltage that can obtained will be equal to the supply voltage w.r.t ground and beyond which the amplifier will saturate

The ideal op-amp has infinite high input resistance and infinite voltage gain.

Open loop voltage gain is given by $A _v=\dfrac{V _0}{V _+-V _-}$. If $V _+=V _-$, Voltage gain becomes infinity or very high. thus voltage gain increases

An ideal op-amp exhibits zero output resistance so that output can drive an infinite number of other devices.

Since input impedance for an ideal opamp is infinite, the current drawn will be zero 

a voltage follower is a circuit, wherein the magnitude and the phase of the signal remains the same. In order to achieve it, the feedback signal should be fed to positive terminal, so that phase is unchanged and the absence of a feedback resistor makes no loss in the output signal

The input impedance of an ideal opamp is infinite and hence the current through it is zero. Thus the voltage across the terminals is also zero

Open loop gain of an amplifier is the ratio of output voltage to the input voltage difference that is being amplified.

The significant AC parameters of Op-amp are transient response, bandwidth and slew rate but all of them have different results while becoming excessive. The transient response affects the settling time while slew rate give rise to distortion if exceeded. As a result, bandwidth reduces the output voltage when it is exceeded. 

Since bandwidth is a small signal phenomenon, it represents the band of frequencies for which the gain remains constant & also highly dependable on compensating components including closed loop gain. But, it exhibits adverse consequences in the reduction of an output voltage when exceeded to greater extent.

Open loop voltage in an opamp is very high and hence option d is the correct option

1. Virtual ground refers to a circuit elements that is not connected to the ground directly, it is held at a reference voltage. This reference voltage need not to be same voltage as ground either.
   2. While in all other ground it may be mechanical A.C or an ordinary ground it act as a reference point against which other voltage can be measured.
   3. The key point  is that the current flow from one point of the circuit through ground then back into the circuit.
   So, current cannot flow to ground through a virtual ground.

Open loop configuration of Op-amp may increase the threat of distortion as well as clipping of output signal. Due to this reason, the output gets switched between the positive and negative saturation levels.

Hence, this is the major drawback because of which it is highly impossible to use Op-amps with open loop configurations in linear circuit applications.

the closed loop gain is given by $A _{cl}=\dfrac{A}{1+A \beta}$. since the open loop gain is large, the closed loop gain almost remains a constant, as it will be independent of open loop gain 

Input current in an ideal operational amplifier   $i = \dfrac{V}{Z}$
Since,  input impedance of the ideal operational amplifier is infinite  $i.e. Z = \infty$
$\implies \ i =\dfrac{V}{\infty}= 0$