In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will be:

We have to find the current that can flow when all appliances in the building are switched on together, because the main fuse must be able to carry at least this much current without melting.

First, let us list every appliance with its power rating and then calculate the total power.

There are 15 bulbs each of 40 W. Their total power is

$$P_1 = 15 \times 40\ \text{W} = 600\ \text{W}.$$

Next, there are 5 bulbs each of 100 W. Their total power is

$$P_2 = 5 \times 100\ \text{W} = 500\ \text{W}.$$

We also have 5 fans, each rated at 80 W. Their total power becomes

$$P_3 = 5 \times 80\ \text{W} = 400\ \text{W}.$$

Finally, there is one heater rated at 1 kW, that is

$$P_4 = 1\ \text{kW} = 1000\ \text{W}.$$

Now we add all these individual powers to obtain the total power consumption when every appliance is on:

$$P_{\text{total}} = P_1 + P_2 + P_3 + P_4 = 600\ \text{W} + 500\ \text{W} + 400\ \text{W} + 1000\ \text{W} = 2500\ \text{W}.$$

The electric mains supply a voltage of 220 V. We use the electric power formula, which states

$$P = VI,$$

where $$P$$ is the power, $$V$$ the voltage and $$I$$ the current.

Rearranging for current, we get

$$I = \frac{P}{V}.$$

Substituting the total power and the mains voltage, we find

$$I_{\text{total}} = \frac{2500\ \text{W}}{220\ \text{V}} = \frac{2500}{220}\ \text{A}.$$

Carrying out the division step by step,

$$\frac{2500}{220} = \frac{25}{2.2} = 11.36\ \text{A} \;(\text{approximately}).$$

So, about 11.36 A of current will flow when everything is working simultaneously. The fuse must have a rating just higher than this value so that it does not blow during normal operation but will still protect the circuit if the current exceeds this level.

Among the given standard fuse ratings in the options — 8 A, 10 A, 12 A and 14 A — the smallest rating that safely exceeds 11.36 A is 12 A.

Hence, the correct answer is Option C.