Given Information:
- Length of the nichrome wire, L = 10 m
- Resistance of the wire, R = 80 Ω
- Current carrying capacity of the wire, I = 5 A
- Voltage of the mains supply, V = 200 V

Calculating Power:
The power (P) is given by the formula:
P = V * I

Substituting the values, we get:
P = 200 V * 5 A
P = 1000 W

Since 1 kW = 1000 W, the power can be expressed as:
P = 1 kW

Dividing the Wire:
The wire can be divided into equal parts and connected either in series or parallel.

Series Connection:
When the wire is connected in series, the total resistance (R_total) is given by the formula:
R_total = R1 + R2 + R3 + ...

Since the wire is divided into equal parts, the resistance of each part (R_part) is:
R_part = R / n

where n is the number of equal parts.

Substituting the given values, we get:
R_part = 80 Ω / n

The resistance of all the parts connected in series is:
R_total = (80 Ω / n) + (80 Ω / n) + (80 Ω / n) + ...

Since the number of parts is not given, let's assume it to be m.

R_total = m * (80 Ω / n)

For maximum power transfer in a series circuit, the load resistance should be equal to the resistance of the circuit. Therefore, for maximum power transfer, R_total = R.

m * (80 Ω / n) = 80 Ω
m/n = 1

This means, the number of parts (m) should be equal to the number of equal parts (n) for maximum power transfer.

Parallel Connection:
When the wire is connected in parallel, the total resistance (R_total) is given by the formula:
1/R_total = 1/R1 + 1/R2 + 1/R3 + ...

Since the wire is divided into equal parts, the resistance of each part (R_part) is:
R_part = R / n

where n is the number of equal parts.

Substituting the given values, we get:
R_part = 80 Ω / n

The resistance of all the parts connected in parallel is:
1/R_total = 1/(80 Ω / n) + 1/(80 Ω / n) + 1/(80 Ω / n) + ...

Simplifying the equation, we get:
1/R_total = n/(80 Ω)

For maximum power transfer in a parallel circuit, the load resistance should be equal to the resistance of the circuit. Therefore, for maximum power transfer, R_total = R.

n/(80 Ω) = 1/80 Ω
n = 1

This means, the number of equal parts (n) should be 1 for maximum power transfer.

Maximum Power:
From the above calculations, we can see that the maximum power can be obtained when the wire is divided into equal parts and connected in parallel, with a single part.

Therefore, the maximum power is 1 kW, which