The masses of the three wires of copper are in the ratio `5:3:1` and their lengths are in the ratio `1:3:5`. The ratio of their electrical resistances is

Options

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

Answer

To find the ratio of electrical resistances of the three wires of copper, we can follow these steps:

### Step 1: Understand the relationship between resistance, length, and area
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 resistances in terms of given ratios
Let the masses of the wires be \( M_1, M_2, M_3 \) in the ratio \( 5:3:1 \), and the lengths be \( L_1, L_2, L_3 \) in the ratio \( 1:3:5 \).

Thus, we can write:
\[ M_1 : M_2 : M_3 = 5 : 3 : 1 \]
\[ L_1 : L_2 : L_3 = 1 : 3 : 5 \]

### Step 3: Relate area to 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:
\[ M = \rho \cdot V \]
Since the density \( \rho \) is constant for copper, we can express the area \( A \) in terms of mass \( M \) and length \( L \):
\[ A = \frac{M}{\rho L} \]

### Step 4: Substitute area in the resistance formula
Substituting \( A \) into the resistance formula gives:
\[ R = \frac{\rho L}{\frac{M}{\rho L}} = \frac{\rho^2 L^2}{M} \]
Thus, the ratio of resistances can be expressed as:
\[ R_1 : R_2 : R_3 = \frac{L_1^2}{M_1} : \frac{L_2^2}{M_2} : \frac{L_3^2}{M_3} \]

### Step 5: Substitute the ratios for lengths and masses
Now we substitute the ratios:
- \( L_1 = 1 \), \( L_2 = 3 \), \( L_3 = 5 \)
- \( M_1 = 5 \), \( M_2 = 3 \), \( M_3 = 1 \)

Thus:
\[ R_1 : R_2 : R_3 = \frac{1^2}{5} : \frac{3^2}{3} : \frac{5^2}{1} \]
Calculating each term:
\[ = \frac{1}{5} : \frac{9}{3} : \frac{25}{1} = \frac{1}{5} : 3 : 25 \]

### Step 6: Simplify the ratio
To express this in integer form, we can multiply each term by 5:
\[ 1 : 15 : 125 \]

### Final Result
Thus, the ratio of the electrical resistances \( R_1 : R_2 : R_3 \) is:
\[ \boxed{1 : 15 : 125} \]