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

Options

A. `1:03:05`
B. `5:03:01`
C. `1 : 15 : 125`
D. `125 : 15 : 1`

Explanation

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

### Step 1: Understand the given ratios
- The masses of the three wires are in the ratio \( M_1 : M_2 : M_3 = 1 : 3 : 5 \).
- The lengths of the three wires are in the ratio \( L_1 : L_2 : L_3 = 5 : 3 : 1 \).

### Step 2: Use the formula for resistance
The resistance \( R \) of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
where:
- \( \rho \) is the resistivity of the material (constant for copper),
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.

### Step 3: Relate volume to mass and density
The volume \( V \) of the wire can be expressed in terms of mass \( M \) and density \( D \):
\[ V = \frac{M}{D} \]
For a cylindrical wire, the cross-sectional area \( A \) can be expressed as:
\[ A = \frac{V}{L} = \frac{M}{D \cdot L} \]

### Step 4: Substitute the area in the resistance formula
Substituting \( A \) into the resistance formula gives:
\[ R = \frac{\rho L}{\frac{M}{D \cdot L}} = \frac{\rho D L^2}{M} \]

### Step 5: Write the expressions for the resistances
Now, we can write the expressions for the resistances of the three wires:
- For wire 1:
\[ R_1 = \frac{\rho D L_1^2}{M_1} \]
- For wire 2:
\[ R_2 = \frac{\rho D L_2^2}{M_2} \]
- For wire 3:
\[ R_3 = \frac{\rho D L_3^2}{M_3} \]

### Step 6: Find the ratios of resistances
Now, we can find the ratios of the resistances:
\[ \frac{R_1}{R_2} = \frac{L_1^2 / M_1}{L_2^2 / M_2} = \frac{L_1^2 \cdot M_2}{L_2^2 \cdot M_1} \]
\[ \frac{R_2}{R_3} = \frac{L_2^2 / M_2}{L_3^2 / M_3} = \frac{L_2^2 \cdot M_3}{L_3^2 \cdot M_2} \]

### Step 7: Substitute the ratios
Using the given ratios:
- \( L_1 : L_2 : L_3 = 5 : 3 : 1 \) implies \( L_1 = 5k, L_2 = 3k, L_3 = 1k \).
- \( M_1 : M_2 : M_3 = 1 : 3 : 5 \) implies \( M_1 = 1m, M_2 = 3m, M_3 = 5m \).

Now substituting these values:
\[ \frac{R_1}{R_2} = \frac{(5k)^2 \cdot (3m)}{(3k)^2 \cdot (1m)} = \frac{25k^2 \cdot 3m}{9k^2 \cdot 1m} = \frac{75}{9} = \frac{25}{3} \]
\[ \frac{R_2}{R_3} = \frac{(3k)^2 \cdot (5m)}{(1k)^2 \cdot (3m)} = \frac{9k^2 \cdot 5m}{1k^2 \cdot 3m} = \frac{45}{3} = 15 \]

### Step 8: Final ratio
Now we can express the final ratio \( R_1 : R_2 : R_3 \):
- From \( \frac{R_1}{R_2} = \frac{25}{3} \) and \( \frac{R_2}{R_3} = 15 \):
\[ R_1 = \frac{25}{3} R_2 \]
\[ R_2 = 15 R_3 \]

Substituting \( R_2 \) in terms of \( R_3 \) into the equation for \( R_1 \):
\[ R_1 = \frac{25}{3} \cdot 15 R_3 = 125 R_3 \]

Thus, the final ratio is:
\[ R_1 : R_2 : R_3 = 125 : 15 : 1 \]

### Final Answer
The ratio of their electrical resistance is \( 125 : 15 : 1 \).