The op-amp and the 1 mA current source in the circuit of figure are ideal. The output of the op-amp is:
When the switch $S_2$ is closed, the gain of the programmable gain amplifier shown in the following figure is:
For the op-amp circuit shown in the figure below, $V_o$ is:
If the op-amp in figure is ideal, the output voltage Vout will be equal to:
Match List-I with List-II 
Match List-I with List-II\[\begin{array}{|c|c|} \hline \textbf{Provision} & \textbf{Case Law} \\ \hline \text{(A) Strict Liability} & \text{(1) Ryland v. Fletcher} \\ \hline \text{(B) Absolute Liability} & \text{(II) M.C. Mehta v. Union of India} \\ \hline \text{(C) Negligence} & \text{(III) Nicholas v. Marsland} \\ \hline \text{(D) Act of God} & \text{(IV) MCD v. Subhagwanti} \\ \hline \end{array}\]