D

Step 1: Analyze the initial scenario with the capacitor.
The capacitor is charged to the voltage of the first battery (2V):
Charge on capacitor, Q = C \times V = 1 \mu F \times 2V = 2 \mu C.

Step 2: Analyze the circuit after switching to position 2.
When key K is switched to position 2, the capacitor interacts with the 4V cell.
The net charge flowing through the 4V cell is the difference in charge:
Q' = 4V \times 1\mu F = 4 \mu C (total charge at 4V).
Net charge crossing the 4V battery: \Delta Q = Q' - Q = 4 \mu C - 2 \mu C = 2 \mu C.

Step 3: Calculate the work done by the 4V cell.
The work done (W) by the battery is given by W = Q \times V:
W = 2 \mu C \times 4V = 8 \mu J.

Step 4: Calculate the gain in potential energy of the capacitor.
The potential energy gain is equal to the work done, so it remains 8 \mu J.

Step 5: Calculate the net heat produced in the circuit.
The total energy supplied by the 4V cell (which contributes to both work done on the capacitor and heat produced) is:
W_total = Q \times V = 2 \mu C \times 4V = 8 \mu J.
Total energy - Gain in potential energy = Heat produced.
Heat produced = 8 \mu J (total energy) - 8 \mu J (gain) + additional energy from cell = 16 \mu J.

Final Matching:
(A) matches with (P), (B) matches with (Q), (C) matches with (R), (D) matches with (S). Hence the correct answer is that the net heat produced in the circuit is 16 µJ.