If the diameter of a conductor is halved, what happens to its resistance?

• A. Resistance remains unchanged
• B. Resistance is doubled
• C. Resistance is quadrupled
• D. Resistance is halved

Answer: (C)

When the diameter of a conductor is halved, the cross-sectional area of the conductor decreases significantly. The resistance of a conductor is calculated using the formula:

[ R = \frac{\rho L}{A} ]

where ( R ) is the resistance, ( \rho ) is the resistivity of the material, ( L ) is the length of the conductor, and ( A ) is the cross-sectional area.

The cross-sectional area of a circular conductor is determined by the formula:

[ A = \pi \left( \frac{d}{2} \right)^2 = \frac{\pi d^2}{4} ]

So, when the diameter ( d ) is halved, the new diameter becomes ( \frac{d}{2} ), and the new area can be calculated as:

[ A' = \pi \left( \frac{d/2}{2} \right)^2 = \pi \left( \frac{d}{4} \right)^2 = \frac{\pi d^2}{16} ]

This shows that the new cross-sectional area is one-fourth of the original area because (\frac{\pi d^2}{16}