Two copper wires have their masses in the ratio 2 : 3 and the lengths in the ratio 3 : 4. The ratio of the resistances is

Options

A. `4:9`
B. `27:32`
C. `16:9`
D. `27:128`

Explanation

To find the ratio of the resistances of two copper wires given their mass and length ratios, we can follow these steps:

### Step 1: Understand the relationship between mass, volume, and density
The mass of a wire can be expressed in terms of its density and volume. The volume of a wire can be expressed as the product of its cross-sectional area and length:
\[ \text{Volume} = \text{Area} \times \text{Length} \quad \Rightarrow \quad V = A \times L \]
Thus, the mass \( m \) can be written as:
\[ m = \text{Density} \times \text{Volume} \quad \Rightarrow \quad m = d \times A \times L \]

### Step 2: Set up the ratios for the two wires
Let the masses of the two wires be \( m_1 \) and \( m_2 \), and their lengths be \( l_1 \) and \( l_2 \). Given:
\[ \frac{m_1}{m_2} = \frac{2}{3} \quad \text{and} \quad \frac{l_1}{l_2} = \frac{3}{4} \]

### Step 3: Express mass in terms of density and volume
For the two wires:
\[ m_1 = d \cdot A_1 \cdot l_1 \quad \text{and} \quad m_2 = d \cdot A_2 \cdot l_2 \]
Substituting these into the mass ratio gives:
\[ \frac{d \cdot A_1 \cdot l_1}{d \cdot A_2 \cdot l_2} = \frac{2}{3} \]
The density \( d \) cancels out:
\[ \frac{A_1 \cdot l_1}{A_2 \cdot l_2} = \frac{2}{3} \]

### Step 4: Substitute the length ratio
Using the length ratio \( \frac{l_1}{l_2} = \frac{3}{4} \):
\[ \frac{A_1 \cdot \frac{3}{4} l_2}{A_2 \cdot l_2} = \frac{2}{3} \]
This simplifies to:
\[ \frac{A_1 \cdot 3}{A_2 \cdot 4} = \frac{2}{3} \]

### Step 5: Solve for the area ratio
Cross-multiplying gives:
\[ 3A_1 \cdot 3 = 2A_2 \cdot 4 \quad \Rightarrow \quad 9A_1 = 8A_2 \quad \Rightarrow \quad \frac{A_1}{A_2} = \frac{8}{9} \]

### Step 6: Calculate the resistance ratio
The resistance \( R \) of a wire is given by:
\[ R = \rho \cdot \frac{l}{A} \]
Thus, the ratio of the resistances \( R_1 \) and \( R_2 \) is:
\[ \frac{R_1}{R_2} = \frac{\rho \cdot l_1 / A_1}{\rho \cdot l_2 / A_2} = \frac{l_1 \cdot A_2}{l_2 \cdot A_1} \]
Substituting the known ratios:
\[ \frac{R_1}{R_2} = \frac{l_1}{l_2} \cdot \frac{A_2}{A_1} = \frac{3/4}{8/9} = \frac{3}{4} \cdot \frac{9}{8} = \frac{27}{32} \]

### Final Result
Thus, the ratio of the resistances is:
\[ \frac{R_1}{R_2} = \frac{27}{32} \]