An electric pressing iron is connected to the mains using an insulated wire. wire becomes very hot. The heat…

Analyze the question: The wire connecting the iron to the mains gets very hot. We want to reduce the heat generated in the wire itself, so we need to find which change lowers that heating.

Concept: Heat produced in a wire is given by Joule heating:

H = I²Rt

The heat depends on the resistance R of the wire. Resistance is given by:

R = ρL / A

where ρ is the resistivity, L is the length and A is the cross-sectional area. Since A = πd²/4, the area depends on the diameter d.

Solution:

1. The iron draws a fixed current, so to reduce H = I²Rt we must reduce R.
2. R = ρL/A, so R falls when A increases.
3. A larger area means a greater diameter, since A = πd²/4.
4. A wire of greater diameter therefore has lower resistance and generates less heat for the same current.

Now check the other options:

• Smaller diameter (C): This raises resistance, so there is more heating. Wrong.
• Thinner insulation (A) and thicker insulation (D): Insulation does not change the resistance of the metal core, so it does not reduce the heat generated. Thicker insulation may even trap heat. Wrong.

Answer: B. greater diameter.

Exam tip: Remember "thick wire, low resistance, less heat". This is why heavy appliances like pressing irons, cookers and heaters need thick cables, while thin wires suit low-current devices. Whenever a question asks about reducing heating in a wire, think R = ρL/A and aim to lower R.