The massses of the three wires of copper are in the ratio 1 : 3 : 5. And their lengths are in th ratio 5 : 3 : 1. the ratio of their electrical resistance is

Options

A. `1:3:5`
B. `5:3:1`
C. `1:15:125`
D. `125:15:1`

Answer

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

### Step 1: Define the Given Ratios
Let the masses of the three wires be:
- Mass of wire 1 (m1) = 1x
- Mass of wire 2 (m2) = 3x
- Mass of wire 3 (m3) = 5x

Let the lengths of the three wires be:
- Length of wire 1 (L1) = 5y
- Length of wire 2 (L2) = 3y
- Length of wire 3 (L3) = 1y

### Step 2: Relate Mass, Volume, and Density
Since the wires are made of copper, they have the same density (ρ). The mass of a wire can be expressed as:
\[ m = \rho \cdot V \]
where V is the volume. The volume of a wire can also be expressed in terms of its cross-sectional area (A) and length (L):
\[ V = A \cdot L \]
Thus, we can write:
\[ m = \rho \cdot A \cdot L \]

### Step 3: Express Volume in Terms of Mass and Length
From the mass expressions:
- For wire 1: \( V1 = \frac{m1}{\rho} = \frac{1x}{\rho} \)
- For wire 2: \( V2 = \frac{m2}{\rho} = \frac{3x}{\rho} \)
- For wire 3: \( V3 = \frac{m3}{\rho} = \frac{5x}{\rho} \)

### Step 4: Find the Cross-Sectional Area
Using the volume and length relationship:
\[ A = \frac{V}{L} \]
We can find the areas:
- For wire 1:
\[ A1 = \frac{V1}{L1} = \frac{\frac{1x}{\rho}}{5y} = \frac{1x}{5\rho y} \]
- For wire 2:
\[ A2 = \frac{V2}{L2} = \frac{\frac{3x}{\rho}}{3y} = \frac{3x}{3\rho y} = \frac{x}{\rho y} \]
- For wire 3:
\[ A3 = \frac{V3}{L3} = \frac{\frac{5x}{\rho}}{1y} = \frac{5x}{\rho y} \]

### Step 5: Calculate the Ratios of Areas
Now we can find the ratios of the areas:
\[ A1 : A2 : A3 = \frac{1x}{5\rho y} : \frac{x}{\rho y} : \frac{5x}{\rho y} \]
To simplify:
- Multiply through by \( 5\rho y \):
\[ A1 : A2 : A3 = 1 : 5 : 25 \]

### Step 6: Use the Resistance Formula
The resistance \( R \) of a wire is given by:
\[ R = \frac{\rho L}{A} \]
Thus, the ratios of resistance will be:
\[ R1 : R2 : R3 = \frac{L1}{A1} : \frac{L2}{A2} : \frac{L3}{A3} \]

### Step 7: Substitute the Lengths and Areas
Substituting the values:
- \( R1 = \frac{5y}{\frac{1x}{5\rho y}} = \frac{5y \cdot 5\rho y}{1x} = \frac{25\rho y^2}{x} \)
- \( R2 = \frac{3y}{\frac{x}{\rho y}} = \frac{3y \cdot \rho y}{x} = \frac{3\rho y^2}{x} \)
- \( R3 = \frac{1y}{\frac{5x}{\rho y}} = \frac{1y \cdot \rho y}{5x} = \frac{\rho y^2}{5x} \)

### Step 8: Calculate the Ratios of Resistance
Now we can find the ratios:
\[ R1 : R2 : R3 = \frac{25\rho y^2}{x} : \frac{3\rho y^2}{x} : \frac{\rho y^2}{5x} \]
To simplify:
- Cancel \( \rho y^2/x \):
\[ R1 : R2 : R3 = 25 : 3 : \frac{1}{5} \]

### Step 9: Normalize the Ratios
To eliminate the fraction, multiply through by 5:
\[ R1 : R2 : R3 = 125 : 15 : 1 \]

### Final Answer
Thus, the ratio of their electrical resistances is:
\[ \boxed{125 : 15 : 1} \]