One kg of copper is drawn into a wire of 1 mm diameter and a wire of 2 mm diameter. The resistance of the two wires will be in the ratio

Options

A. `2:1`
B. `1:2`
C. `16:1`
D. `4:1`

Explanation

To find the ratio of the resistances of two copper wires with different diameters but the same mass, we can follow these steps:

### Step 1: Write down the given data
- Mass of copper (m) = 1 kg
- Diameter of wire 1 (D1) = 1 mm = 0.001 m
- Diameter of wire 2 (D2) = 2 mm = 0.002 m

### Step 2: Calculate the cross-sectional area of each wire
The cross-sectional area (A) of a wire with diameter D is given by the formula:
\[ A = \frac{\pi D^2}{4} \]

For wire 1:
\[ A_1 = \frac{\pi (D_1)^2}{4} = \frac{\pi (0.001)^2}{4} = \frac{\pi (0.000001)}{4} = \frac{\pi}{4000000} \, \text{m}^2 \]

For wire 2:
\[ A_2 = \frac{\pi (D_2)^2}{4} = \frac{\pi (0.002)^2}{4} = \frac{\pi (0.000004)}{4} = \frac{\pi}{1000000} \, \text{m}^2 \]

### Step 3: Use the formula for resistance
The resistance (R) of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
where:
- \( \rho \) = resistivity of the material (constant for copper)
- \( L \) = length of the wire
- \( A \) = cross-sectional area

### Step 4: Relate length and volume
The volume (V) of the wire can be expressed as:
\[ V = A \times L \]
Since the mass of copper is constant, we can express the volume in terms of mass and density:
\[ V = \frac{m}{\rho} \]
Thus, we can express the length in terms of the area:
\[ L = \frac{V}{A} = \frac{m/\rho}{A} \]

### Step 5: Substitute length into the resistance formula
Substituting \( L \) into the resistance formula gives:
\[ R = \frac{\rho \left(\frac{m}{\rho A}\right)}{A} = \frac{m}{\rho A^2} \]
This shows that the resistance is inversely proportional to the square of the area:
\[ R \propto \frac{1}{A^2} \]

### Step 6: Relate area to diameter
Since the area \( A \) is proportional to the square of the diameter:
\[ A \propto D^2 \]
Thus:
\[ R \propto \frac{1}{(D^2)^2} = \frac{1}{D^4} \]

### Step 7: Set up the ratio of resistances
Using the relationship derived:
\[ \frac{R_1}{R_2} = \left(\frac{D_2}{D_1}\right)^4 \]
Substituting the diameters:
\[ \frac{R_1}{R_2} = \left(\frac{2 \, \text{mm}}{1 \, \text{mm}}\right)^4 = \left(2\right)^4 = 16 \]

### Final Answer
The ratio of the resistances \( R_1 : R_2 \) is \( 16 : 1 \).

---