Question

When a wire is stretched and its radius becomes R by 2 then its resistance will be?

Answer 1

To determine how the resistance of a wire changes when its radius is altered, we can use the formula for the resistance of a wire:

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

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

The cross-sectional area \( A \) of a wire with radius \( r \) is given by:

\[ A = \pi r^2 \]

If the radius of the wire is reduced to \( \frac{R}{2} \), the new cross-sectional area \( A' \) is:

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

Since the cross-sectional area is reduced to a quarter of the original area, the resistance \( R' \) of the wire will be:

\[ R' = \rho \frac{L}{A'} = \rho \frac{L}{\frac{A}{4}} = 4 \cdot \rho \frac{L}{A} = 4R \]

So, the resistance of the wire will increase by a factor of 4 when its radius is halved.

Answer 2

To understand how the resistance of a wire changes when its radius is altered, we need to consider how resistance depends on the dimensions of the wire.

**Resistance of a Wire:**

The resistance \( R \) of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
where:
- \( \rho \) is the resistivity of the material (a property that does not change with the wire's dimensions),
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.

For a wire with a circular cross-section, the area \( A \) can be expressed in terms of its radius \( r \):
\[ A = \pi r^2 \]

**Change in Radius:**

Let's denote the original radius of the wire as \( r \). If the wire is stretched such that its radius becomes \( \frac{r}{2} \), the new cross-sectional area \( A' \) of the wire will be:
\[ A' = \pi \left(\frac{r}{2}\right)^2 = \pi \frac{r^2}{4} \]

So, the new area is \( \frac{1}{4} \) of the original area.

**Resistance Calculation:**

1. **Original Resistance:**
   \[ R = \frac{\rho L}{A} = \frac{\rho L}{\pi r^2} \]

2. **New Resistance with Reduced Radius:**
   \[ R' = \frac{\rho L}{A'} = \frac{\rho L}{\pi \frac{r^2}{4}} = \frac{4 \rho L}{\pi r^2} \]

Comparing this with the original resistance:
   \[ R' = 4 R \]

**Conclusion:**

When the radius of the wire is reduced to half of its original value, its cross-sectional area becomes one-fourth of the original. As a result, the resistance of the wire increases by a factor of 4. Therefore, the new resistance will be four times the original resistance.