Given data:
- Transformer rating: 10 kVA
- Voltage ratio: 500/250 V
- S.C. test results: 60 V, 20 A, 150 W
- Maximum efficiency occurs at unity pf and at 1.2 full load current

To solve the problem, we need to use the following formulas and concepts:

- Efficiency (η) = Output power / Input power
- Output power = Voltage x Current
- Input power = Voltage x Current + Copper losses + Iron losses
- Copper losses = I²R losses in windings
- Iron losses = Hysteresis losses + Eddy current losses

(i) Finding the maximum efficiency:
To find the maximum efficiency, we need to plot the efficiency curve of the transformer. The efficiency curve is a plot of efficiency vs. load current at different power factors. We can use the following steps to plot the efficiency curve:

1. Calculate the copper losses:
- From the S.C. test, we know that the short-circuit current (Isc) is 20 A at 60 V.
- Since the transformer is single-phase, the copper losses (Pcu) are equal to Isc² x R, where R is the resistance of the winding.
- We don't know the resistance of the winding, but we can calculate it using the voltage drop in the winding at full load.
- At full load, the voltage drop in the winding is (500 - 250) = 250 V.
- Therefore, the resistance of the winding is R = 250 / 20 = 12.5 Ω.
- Therefore, the copper losses are Pcu = Isc² x R = 20² x 12.5 = 5000 W.

2. Calculate the iron losses:
- The iron losses (Pi) are the sum of hysteresis losses and eddy current losses.
- We don't know the values of hysteresis losses and eddy current losses, but we can estimate them using the rated voltage and frequency of the transformer.
- Let's assume that the hysteresis losses and eddy current losses are 2% of the rated power at full load.
- Therefore, the iron losses are Pi = 0.02 x 10,000 = 200 W.

3. Calculate the output power:
- The output power (Pout) at full load is equal to the rated power, which is 10 kVA (or 10,000 W).
- At 1.2 full load current, the output voltage is 250 V (since the voltage ratio is 500/250), and the output current is 1.2 x 10,000 / 250 = 480 A.
- Therefore, the output power at 1.2 full load current and unity pf is Pout = 250 x 480 = 120,000 W.

4. Calculate the input power:
- At 1.2 full load current and unity pf, the input voltage is 500 V, and the input current is 480 / 0.8 = 600 A (since the power factor is unity).
- Therefore, the input power is Pinput = 500 x 600 + 5000 + 200 = 305,200 W.

5. Calculate the efficiency:
- The efficiency at 1.2 full load current and unity pf is