If the radius of cross-section of the conductor is increased by `0.1 %` keeping volume constant, then percentage change in the resistance of the conductor is

Options

A. `- 0.2 %`
B. ` - 0.1 %`
C. `-0.3 %`
D. `-0.4%`

Answer

To solve the problem of finding the percentage change in the resistance of a conductor when the radius of its cross-section is increased by 0.1% while keeping the volume constant, we can follow these steps:

### Step 1: Understand the relationship between radius, area, and volume
The volume \( V \) of a cylindrical conductor is given by:
\[ V = A \times L \]
where \( A \) is the cross-sectional area and \( L \) is the length of the conductor. The area \( A \) of a circle is given by:
\[ A = \pi r^2 \]
where \( r \) is the radius.

### Step 2: Define initial and final conditions
Let:
- Initial radius \( r_1 \)
- New radius \( r_2 = r_1 + 0.1\% \text{ of } r_1 = r_1 + 0.001 \times r_1 = 1.001 r_1 \)

### Step 3: Set up the volume equation
Since the volume is constant, we can write:
\[ L_1 \times A_1 = L_2 \times A_2 \]
Substituting the areas:
\[ L_1 \times \pi r_1^2 = L_2 \times \pi r_2^2 \]
Cancelling \( \pi \) gives:
\[ L_1 r_1^2 = L_2 r_2^2 \]

### Step 4: Substitute for \( r_2 \)
Substituting \( r_2 = 1.001 r_1 \):
\[ L_1 r_1^2 = L_2 (1.001 r_1)^2 \]
This simplifies to:
\[ L_1 r_1^2 = L_2 (1.002001 r_1^2) \]
Cancelling \( r_1^2 \):
\[ L_1 = L_2 \times 1.002001 \]
Thus,
\[ L_2 = \frac{L_1}{1.002001} \]

### Step 5: Calculate initial and new resistance
The resistance \( R \) of a conductor is given by:
\[ R = \frac{\rho L}{A} \]
where \( \rho \) is the resistivity.

- Initial resistance \( R_1 \):
\[ R_1 = \frac{\rho L_1}{\pi r_1^2} \]

- New resistance \( R_2 \):
\[ R_2 = \frac{\rho L_2}{\pi r_2^2} \]
Substituting \( L_2 \) and \( r_2 \):
\[ R_2 = \frac{\rho \left(\frac{L_1}{1.002001}\right)}{\pi (1.001 r_1)^2} = \frac{\rho L_1}{\pi r_1^2} \cdot \frac{1}{1.002001 \cdot 1.002001} = R_1 \cdot \frac{1}{1.002001^2} \]

### Step 6: Find the percentage change in resistance
The percentage change in resistance is given by:
\[ \text{Percentage Change} = \frac{R_2 - R_1}{R_1} \times 100 \]
Substituting for \( R_2 \):
\[ \text{Percentage Change} = \left(\frac{R_1 \cdot \frac{1}{1.002001^2} - R_1}{R_1}\right) \times 100 = \left(\frac{1 - \frac{1}{1.002001^2}}{1}\right) \times 100 \]

Calculating \( 1.002001^2 \):
\[ 1.002001^2 \approx 1.004004 \]
Thus,
\[ \text{Percentage Change} \approx \left(1 - \frac{1}{1.004004}\right) \times 100 \approx \left(1 - 0.996015\right) \times 100 \approx 0.3985\% \]
This is approximately \( -0.4\% \).

### Final Answer
The percentage change in the resistance of the conductor is approximately:
\[ \text{Percentage Change} \approx -0.4\% \]