Question

What happens to the resistance of a wire if its radius is halved?

• A. It becomes half
• B. It becomes double
• C. It increases 4 times
• D. It increases 16 times

Hint:

Resistance is inversely proportional to the square of the radius. Small changes in radius cause large changes in resistance. If radius is halved, resistance becomes \(4\) times; if radius is doubled, resistance becomes \(\frac{1}{4}\).

Answer 1

The Correct Option is C

Concept:
Resistance (\(R\)) of a conductor depends on its length (\(L\)), cross-sectional area (\(A\)), and the nature of the material (resistivity, \(\rho\)).
Step 1: Understanding the Question:
We are changing the radius of a cylindrical wire to half its original value (\(r_{new} = r/2\)) and need to calculate how this affects the electrical resistance.
Step 2: Key Formula or Approach:
Use the resistance formula:
\[ R = \rho \frac{L}{A} \]
For a wire with radius \(r\), the area \(A = \pi r^2\).
Therefore, \(R = \frac{\rho L}{\pi r^2}\).
This implies \(R \propto \frac{1}{r^2}\).
Step 3: Detailed Solution:
Let original resistance be \(R_1 = \frac{k}{r_1^2}\).
The new radius \(r_2 = \frac{r_1}{2}\).
The new resistance \(R_2\) is:
\[ R_2 = \frac{k}{(r_1/2)^2} = \frac{k}{r_1^2/4} \]
\[ R_2 = 4 \times \frac{k}{r_1^2} \]
\[ R_2 = 4 R_1 \]
The resistance increases by a factor of 4.
Step 4: Final Answer:
The resistance increases 4 times.