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

Options

A. `1:2:3`
B. `1:3:5`
C. `125: 15:1`
D. `1:25:125`

Answer

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

### Step 1: Understand the relationship between resistance, length, and area
The electrical resistance \( R \) of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
where:
- \( R \) is the resistance,
- \( \rho \) 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 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 density \( d \) of the material is given by:
\[ d = \frac{m}{V} \]
where \( m \) is the mass of the wire. Rearranging this gives:
\[ A = \frac{m}{d \cdot L} \]

### Step 3: Substitute the area into the resistance formula
Substituting \( A \) into the resistance formula gives:
\[ R = \frac{\rho L}{\frac{m}{d \cdot L}} = \frac{\rho L^2 d}{m} \]

### Step 4: Set up the ratios for the three wires
Let the masses of the wires be \( m_1, m_2, m_3 \) in the ratio \( 1:3:5 \) and their lengths be \( L_1, L_2, L_3 \) in the ratio \( 5:3:1 \).

Thus, we can express:
- \( m_1 = k \)
- \( m_2 = 3k \)
- \( m_3 = 5k \)

And for lengths:
- \( L_1 = 5l \)
- \( L_2 = 3l \)
- \( L_3 = l \)

### Step 5: Calculate the resistance for each wire
Now we can express the resistance for each wire:
\[ R_1 = \frac{\rho (5l)^2 d}{k} = \frac{25\rho l^2 d}{k} \]
\[ R_2 = \frac{\rho (3l)^2 d}{3k} = \frac{9\rho l^2 d}{3k} = \frac{3\rho l^2 d}{k} \]
\[ R_3 = \frac{\rho (l)^2 d}{5k} = \frac{\rho l^2 d}{5k} \]

### Step 6: Find the ratio of resistances
Now we can find the ratio \( R_1 : R_2 : R_3 \):
\[ R_1 : R_2 : R_3 = \frac{25\rho l^2 d}{k} : \frac{3\rho l^2 d}{k} : \frac{\rho l^2 d}{5k} \]
This simplifies to:
\[ R_1 : R_2 : R_3 = 25 : 3 : \frac{1}{5} \]

To make it easier to compare, we can multiply through by 5 to eliminate the fraction:
\[ R_1 : R_2 : R_3 = 125 : 15 : 1 \]

### Final Ratio
Thus, the ratio of their electrical resistances is:
\[ R_1 : R_2 : R_3 = 125 : 15 : 1 \]