It is desired to make a long cylindrical conductor whose temperature coefficient of resistivity at `20^(@)C` will be close to zero. If such a conductor is made by assembling alternate disks of iron and carbon, find the ratio of the thickness of a carbon disk to that an iron disk. `("For carbon", p = 3500 xx 10^(-8)Omega m and alpha= -0.50 xx 10^(-3) .^(@)C^(-1) "for iron, p"=9.68 xx 10^(-8)Omega m and alpha=6.5 xx 10^(-3).^(@)C^(-1))`

Options

A. 0.36
B. 0.036
C. 1
D. 2

Explanation

To solve the problem of finding the ratio of the thickness of a carbon disk to that of an iron disk in a long cylindrical conductor, we will follow these steps:

### Step 1: Understand the Problem
We need to create a cylindrical conductor made of alternating disks of iron and carbon such that the overall temperature coefficient of resistivity (α) at 20°C is zero.

### Step 2: Write the Resistance Formula
The resistance \( R \) of a conductor is given by:
\[ R = \frac{\rho L}{A} \]
where \( \rho \) is the resistivity, \( L \) is the length (or thickness of the disk), and \( A \) is the cross-sectional area.

### Step 3: Consider the Temperature Dependence of Resistance
The resistance changes with temperature according to:
\[ R = R_0 (1 + \alpha (T - T_0)) \]
where \( R_0 \) is the resistance at a reference temperature \( T_0 \), \( \alpha \) is the temperature coefficient of resistivity, and \( T \) is the temperature.

### Step 4: Set Up the Equation for Series Resistance
Since the disks are in series, the total resistance \( R \) is:
\[ R = R_C + R_I \]
where \( R_C \) is the resistance of the carbon disk and \( R_I \) is the resistance of the iron disk.

### Step 5: Express the Resistances in Terms of Resistivity
For carbon:
\[ R_C = \frac{\rho_C L_C}{A} \]
For iron:
\[ R_I = \frac{\rho_I L_I}{A} \]
Thus, the total resistance becomes:
\[ R = \frac{\rho_C L_C}{A} + \frac{\rho_I L_I}{A} \]

### Step 6: Substitute into the Temperature Dependence Equation
Substituting for \( R_C \) and \( R_I \) in the temperature dependence equation gives:
\[ \frac{\rho_C L_C}{A} + \frac{\rho_I L_I}{A} = R_0 \left(1 + \alpha_C (T - T_0)\right) + R_0 \left(1 + \alpha_I (T - T_0)\right) \]

### Step 7: Group Terms
Rearranging gives:
\[ \frac{\rho_C L_C + \rho_I L_I}{A} = R_0 \left(2 + (\alpha_C + \alpha_I)(T - T_0)\right) \]

### Step 8: Set the Coefficient of \( (T - T_0) \) to Zero
For the overall temperature coefficient to be zero, we need:
\[ \rho_C L_C \alpha_C + \rho_I L_I \alpha_I = 0 \]

### Step 9: Solve for the Ratio of Thicknesses
Rearranging gives:
\[ \frac{L_C}{L_I} = -\frac{\rho_I \alpha_I}{\rho_C \alpha_C} \]

### Step 10: Substitute the Given Values
Given:
- For carbon: \( \rho_C = 3500 \times 10^{-8} \, \Omega \cdot m \), \( \alpha_C = -0.50 \times 10^{-3} \, ^{\circ}C^{-1} \)
- For iron: \( \rho_I = 9.68 \times 10^{-8} \, \Omega \cdot m \), \( \alpha_I = 6.5 \times 10^{-3} \, ^{\circ}C^{-1} \)

Substituting these values:
\[ \frac{L_C}{L_I} = -\frac{(9.68 \times 10^{-8})(6.5 \times 10^{-3})}{(3500 \times 10^{-8})(-0.50 \times 10^{-3})} \]

### Step 11: Calculate the Ratio
Calculating the above expression:
\[ \frac{L_C}{L_I} = \frac{9.68 \times 6.5}{3500 \times 0.50} = \frac{62.92}{1750} \approx 0.036 \]

### Final Answer
The ratio of the thickness of a carbon disk to that of an iron disk is:
\[ \frac{L_C}{L_I} \approx 0.036 \]