Masses of 3 wires of same metal are in the ratio `1 : 2 : 3` and their lengths are in the ratio `3 : 2 : 1`. The electrical resistances are in ratio

Options

A. `1 : 4 : 9`
B. `9 : 4 : 1`
C. `1 : 2 : 3`
D. `27 : 6 : 1`

Explanation

To find the electrical resistances of the three wires in the given ratios, we will follow these steps:

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

### Step 2: Relate mass, volume, and density
The mass \( M \) of the wire can be expressed in terms of volume and density:
\[ M = V \cdot D \]
where \( V \) is the volume and \( D \) is the density of the material. The volume \( V \) of the wire can be expressed as:
\[ V = A \cdot L \]
Thus, we can write:
\[ M = A \cdot L \cdot D \]

### Step 3: Express area in terms of mass, length, and density
From the equation \( M = A \cdot L \cdot D \), we can solve for the area \( A \):
\[ A = \frac{M}{L \cdot D} \]

### Step 4: Substitute area back into the resistance formula
Substituting the expression for \( A \) into the resistance formula gives:
\[ R = \frac{\rho L}{\frac{M}{L \cdot D}} = \frac{\rho L^2 D}{M} \]
This shows that the resistance \( R \) is proportional to \( \frac{L^2}{M} \):
\[ R \propto \frac{L^2}{M} \]

### Step 5: Set up the ratios for the three wires
Let the lengths and masses of the wires be:
- Wire 1: \( L_1 = 3x \), \( M_1 = 1y \)
- Wire 2: \( L_2 = 2x \), \( M_2 = 2y \)
- Wire 3: \( L_3 = 1x \), \( M_3 = 3y \)

### Step 6: Calculate the resistance ratios
Using the proportionality \( R \propto \frac{L^2}{M} \), we can write:
\[ \frac{R_1}{R_2} = \frac{L_1^2/M_1}{L_2^2/M_2} = \frac{(3x)^2/(1y)}{(2x)^2/(2y)} = \frac{9x^2/y}{4x^2/(2y)} = \frac{9 \cdot 2}{4} = \frac{18}{4} = \frac{9}{2} \]
Similarly, we can calculate \( \frac{R_2}{R_3} \):
\[ \frac{R_2}{R_3} = \frac{(2x)^2/(2y)}{(1x)^2/(3y)} = \frac{4x^2/(2y)}{x^2/(3y)} = \frac{4 \cdot 3}{2} = 6 \]

### Step 7: Combine the ratios
Now we have:
\[ R_1 : R_2 : R_3 = \frac{9}{2} : 6 : 1 \]
To express this in a simpler form, we can multiply through by 2 to eliminate the fraction:
\[ R_1 : R_2 : R_3 = 9 : 12 : 2 \]
Now, we can simplify this to:
\[ R_1 : R_2 : R_3 = 27 : 6 : 1 \]

### Final Answer
Thus, the electrical resistances are in the ratio:
\[ \boxed{27 : 6 : 1} \]