The masses of three copper wires are in the ratio 2:3:5 and their lengths are in the ratio 5:3:2. Then, the ratio of their electrical resistance is

Options

A. `1:9:15`
B. `2:3:5`
C. `5:3:2`
D. `125:30:8`

Answer

To find the ratio of electrical resistance of three copper wires given their masses and lengths, we can follow these steps:

### Step 1: Understand the formula for resistance
The resistance \( R \) of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
where:
- \( R \) = resistance
- \( \rho \) = resistivity of the material (constant for copper)
- \( L \) = length of the wire
- \( A \) = cross-sectional area of the wire

### Step 2: Express the cross-sectional area in terms of mass and length
The volume \( V \) of the wire can be expressed as:
\[ V = A \cdot L \]
The mass \( m \) of the wire can be expressed as:
\[ m = \rho_{density} \cdot V = \rho_{density} \cdot (A \cdot L) \]
From this, we can express the area \( A \) as:
\[ A = \frac{m}{\rho_{density} \cdot L} \]

### Step 3: Substitute the area into the resistance formula
Substituting \( A \) into the resistance formula gives:
\[ R = \frac{\rho L}{\frac{m}{\rho_{density} \cdot L}} = \frac{\rho_{density} \cdot L^2}{m} \]
Thus, we can see that:
\[ R \propto \frac{L^2}{m} \]

### Step 4: Determine the ratios of lengths and masses
Given the ratios:
- Masses: \( m_1:m_2:m_3 = 2:3:5 \)
- Lengths: \( L_1:L_2:L_3 = 5:3:2 \)

### Step 5: Calculate the resistance ratios
Using the derived formula \( R \propto \frac{L^2}{m} \):
- For wire 1:
\[ R_1 \propto \frac{(5)^2}{2} = \frac{25}{2} \]
- For wire 2:
\[ R_2 \propto \frac{(3)^2}{3} = \frac{9}{3} = 3 \]
- For wire 3:
\[ R_3 \propto \frac{(2)^2}{5} = \frac{4}{5} \]

### Step 6: Write the resistance ratio
Now we can express the resistance ratio \( R_1:R_2:R_3 \):
\[ R_1:R_2:R_3 = \frac{25/2}{1} : \frac{3}{1} : \frac{4/5}{1} \]
To make calculations easier, we can multiply each term by 10 (to eliminate the fractions):
\[ R_1:R_2:R_3 = \frac{25 \times 10}{2} : 3 \times 10 : \frac{4 \times 10}{5} \]
This simplifies to:
\[ R_1:R_2:R_3 = 125 : 30 : 8 \]

### Step 7: Simplify the ratio
To simplify the ratio \( 125:30:8 \), we can divide each term by their greatest common divisor (GCD). The GCD here is 1, so the ratio remains:
\[ R_1:R_2:R_3 = 125:30:8 \]

### Final Result
Thus, the ratio of their electrical resistance is:
\[ \boxed{125:30:8} \]