When a wire of resistance R is stretched so that its radius becomes half?
2 Answers
When a wire of resistance \( R \) is stretched, the change in its dimensions affects its resistance. Let's analyze this problem step by step:
### 1. Resistance and Wire Dimensions
The resistance \( R \) of a wire is given by the formula:
\[
R = \rho \frac{L}{A}
\]
Where:
- \( \rho \) is the resistivity of the material (constant for a given material),
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.
The cross-sectional area \( A \) for a wire with radius \( r \) is:
\[
A = \pi r^2
\]
### 2. Effect of Stretching on Radius
When the wire is stretched, its volume remains constant, assuming it is a homogeneous material. The volume \( V \) of the wire is given by:
\[
V = A \cdot L = \pi r^2 L
\]
Now, if the radius of the wire becomes half, the new radius \( r' \) will be:
\[
r' = \frac{r}{2}
\]
So the new cross-sectional area \( A' \) becomes:
\[
A' = \pi (r')^2 = \pi \left(\frac{r}{2}\right)^2 = \frac{\pi r^2}{4} = \frac{A}{4}
\]
### 3. Effect of Stretching on Length
Since the volume remains constant, the new length \( L' \) of the wire can be found from the relation:
\[
V = A \cdot L = A' \cdot L'
\]
\[
\pi r^2 L = \frac{\pi r^2}{4} L'
\]
Solving for \( L' \):
\[
L' = 4L
\]
### 4. New Resistance
The new resistance \( R' \) of the wire, after it is stretched, is:
\[
R' = \rho \frac{L'}{A'}
\]
Substitute the new length and area:
\[
R' = \rho \frac{4L}{\frac{A}{4}} = \rho \frac{4L \cdot 4}{A} = 16 \rho \frac{L}{A}
\]
Thus, the new resistance \( R' \) is:
\[
R' = 16R
\]
### Conclusion
When the wire is stretched so that its radius becomes half, the resistance increases by a factor of 16. Therefore, the new resistance is:
\[
R' = 16R
\]
### 1. Resistance and Wire Dimensions
The resistance \( R \) of a wire is given by the formula:
\[
R = \rho \frac{L}{A}
\]
Where:
- \( \rho \) is the resistivity of the material (constant for a given material),
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.
The cross-sectional area \( A \) for a wire with radius \( r \) is:
\[
A = \pi r^2
\]
### 2. Effect of Stretching on Radius
When the wire is stretched, its volume remains constant, assuming it is a homogeneous material. The volume \( V \) of the wire is given by:
\[
V = A \cdot L = \pi r^2 L
\]
Now, if the radius of the wire becomes half, the new radius \( r' \) will be:
\[
r' = \frac{r}{2}
\]
So the new cross-sectional area \( A' \) becomes:
\[
A' = \pi (r')^2 = \pi \left(\frac{r}{2}\right)^2 = \frac{\pi r^2}{4} = \frac{A}{4}
\]
### 3. Effect of Stretching on Length
Since the volume remains constant, the new length \( L' \) of the wire can be found from the relation:
\[
V = A \cdot L = A' \cdot L'
\]
\[
\pi r^2 L = \frac{\pi r^2}{4} L'
\]
Solving for \( L' \):
\[
L' = 4L
\]
### 4. New Resistance
The new resistance \( R' \) of the wire, after it is stretched, is:
\[
R' = \rho \frac{L'}{A'}
\]
Substitute the new length and area:
\[
R' = \rho \frac{4L}{\frac{A}{4}} = \rho \frac{4L \cdot 4}{A} = 16 \rho \frac{L}{A}
\]
Thus, the new resistance \( R' \) is:
\[
R' = 16R
\]
### Conclusion
When the wire is stretched so that its radius becomes half, the resistance increases by a factor of 16. Therefore, the new resistance is:
\[
R' = 16R
\]
When a wire of resistance \( R \) is stretched such that its radius becomes half, several factors contribute to the change in its resistance. To understand this, we need to explore the relationship between resistance, length, cross-sectional area, and resistivity of the wire.
### Key Concepts:
1. **Resistance Formula**: The resistance \( R \) of a wire is given by the formula:
\[
R = \rho \frac{L}{A}
\]
Where:
- \( \rho \) is the resistivity (material property of the wire, constant if material is unchanged).
- \( L \) is the length of the wire.
- \( A \) is the cross-sectional area of the wire.
2. **Cross-sectional Area**: For a cylindrical wire, the cross-sectional area \( A \) is related to its radius \( r \) as:
\[
A = \pi r^2
\]
3. **Volume Conservation**: When the wire is stretched, its volume remains the same because the amount of material doesn't change. The volume of the wire can be expressed as:
\[
V = A \cdot L
\]
Since volume remains constant, we have:
\[
A_1 L_1 = A_2 L_2
\]
Where \( A_1 \) and \( L_1 \) are the original area and length, and \( A_2 \) and \( L_2 \) are the area and length after stretching.
### Step-by-Step Analysis:
1. **Initial State**:
- The initial resistance is \( R_1 = \rho \frac{L_1}{A_1} \).
- Initial cross-sectional area \( A_1 = \pi r_1^2 \), where \( r_1 \) is the original radius.
2. **After Stretching (Radius is Halved)**:
- The new radius \( r_2 = \frac{r_1}{2} \).
- The new cross-sectional area \( A_2 \) becomes:
\[
A_2 = \pi r_2^2 = \pi \left(\frac{r_1}{2}\right)^2 = \frac{\pi r_1^2}{4} = \frac{A_1}{4}
\]
3. **Conservation of Volume**:
- Since the volume remains constant, \( A_1 L_1 = A_2 L_2 \).
- Substituting \( A_2 = \frac{A_1}{4} \), we get:
\[
A_1 L_1 = \frac{A_1}{4} L_2
\]
- Solving for \( L_2 \):
\[
L_2 = 4 L_1
\]
- Thus, the length of the wire becomes four times its original length.
4. **New Resistance**:
- The new resistance \( R_2 \) is:
\[
R_2 = \rho \frac{L_2}{A_2}
\]
- Substituting \( L_2 = 4 L_1 \) and \( A_2 = \frac{A_1}{4} \):
\[
R_2 = \rho \frac{4 L_1}{\frac{A_1}{4}} = \rho \frac{4 L_1 \cdot 4}{A_1} = 16 \rho \frac{L_1}{A_1}
\]
- Hence, \( R_2 = 16 R_1 \).
### Conclusion:
When the wire is stretched so that its radius becomes half, its new resistance becomes **16 times** the original resistance. This is due to both the reduction in cross-sectional area and the increase in length.
### Key Concepts:
1. **Resistance Formula**: The resistance \( R \) of a wire is given by the formula:
\[
R = \rho \frac{L}{A}
\]
Where:
- \( \rho \) is the resistivity (material property of the wire, constant if material is unchanged).
- \( L \) is the length of the wire.
- \( A \) is the cross-sectional area of the wire.
2. **Cross-sectional Area**: For a cylindrical wire, the cross-sectional area \( A \) is related to its radius \( r \) as:
\[
A = \pi r^2
\]
3. **Volume Conservation**: When the wire is stretched, its volume remains the same because the amount of material doesn't change. The volume of the wire can be expressed as:
\[
V = A \cdot L
\]
Since volume remains constant, we have:
\[
A_1 L_1 = A_2 L_2
\]
Where \( A_1 \) and \( L_1 \) are the original area and length, and \( A_2 \) and \( L_2 \) are the area and length after stretching.
### Step-by-Step Analysis:
1. **Initial State**:
- The initial resistance is \( R_1 = \rho \frac{L_1}{A_1} \).
- Initial cross-sectional area \( A_1 = \pi r_1^2 \), where \( r_1 \) is the original radius.
2. **After Stretching (Radius is Halved)**:
- The new radius \( r_2 = \frac{r_1}{2} \).
- The new cross-sectional area \( A_2 \) becomes:
\[
A_2 = \pi r_2^2 = \pi \left(\frac{r_1}{2}\right)^2 = \frac{\pi r_1^2}{4} = \frac{A_1}{4}
\]
3. **Conservation of Volume**:
- Since the volume remains constant, \( A_1 L_1 = A_2 L_2 \).
- Substituting \( A_2 = \frac{A_1}{4} \), we get:
\[
A_1 L_1 = \frac{A_1}{4} L_2
\]
- Solving for \( L_2 \):
\[
L_2 = 4 L_1
\]
- Thus, the length of the wire becomes four times its original length.
4. **New Resistance**:
- The new resistance \( R_2 \) is:
\[
R_2 = \rho \frac{L_2}{A_2}
\]
- Substituting \( L_2 = 4 L_1 \) and \( A_2 = \frac{A_1}{4} \):
\[
R_2 = \rho \frac{4 L_1}{\frac{A_1}{4}} = \rho \frac{4 L_1 \cdot 4}{A_1} = 16 \rho \frac{L_1}{A_1}
\]
- Hence, \( R_2 = 16 R_1 \).
### Conclusion:
When the wire is stretched so that its radius becomes half, its new resistance becomes **16 times** the original resistance. This is due to both the reduction in cross-sectional area and the increase in length.