Four wires of the same diameter are connected, in turn, between two points maintained at a constant potential difference, Their resistivities and lengths are, `rho and L ("wire 1") 1.2rho and 1.2L("wire 2"), 0.9 rho and 0.9L("wire 3") and rho and 1.5L ("wire 4")`. Rank the wires according to hte rates at which energy is dissipated as heat, greatest first,

Options

A. `4 gt 3 gt 1 gt 2 `
B. `4 gt 2 gt 1 gt 3 `
C. `1 gt 2 gt 3 gt 4`
D. `3 gt 1 gt 2 gt 4`

Explanation

To determine the rates at which energy is dissipated as heat in the four wires, we will use the formula for power (P) dissipated in a resistor, which is given by:

\[ P = \frac{V^2}{R} \]

Where:
- \( P \) is the power (rate of energy dissipation),
- \( V \) is the potential difference across the wire,
- \( R \) is the resistance of the wire.

The resistance \( R \) of a wire can be calculated using the formula:

\[ R = \frac{\rho L}{A} \]

Where:
- \( \rho \) is the resistivity of the material,
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.

Since all wires have the same diameter, their cross-sectional area \( A \) is constant and can be factored out when comparing the power dissipated in each wire.

### Step 1: Calculate the resistance for each wire

1. **Wire 1:**
- Resistivity: \( \rho \)
- Length: \( L \)
- Resistance:
\[ R_1 = \frac{\rho L}{A} \]

2. **Wire 2:**
- Resistivity: \( 1.2\rho \)
- Length: \( 1.2L \)
- Resistance:
\[ R_2 = \frac{1.2\rho \cdot 1.2L}{A} = \frac{1.44\rho L}{A} \]

3. **Wire 3:**
- Resistivity: \( 0.9\rho \)
- Length: \( 0.9L \)
- Resistance:
\[ R_3 = \frac{0.9\rho \cdot 0.9L}{A} = \frac{0.81\rho L}{A} \]

4. **Wire 4:**
- Resistivity: \( \rho \)
- Length: \( 1.5L \)
- Resistance:
\[ R_4 = \frac{\rho \cdot 1.5L}{A} = \frac{1.5\rho L}{A} \]

### Step 2: Calculate the power dissipated in each wire

Using the power formula \( P = \frac{V^2}{R} \):

1. **Power for Wire 1:**
\[ P_1 = \frac{V^2}{R_1} = \frac{V^2 A}{\rho L} \]

2. **Power for Wire 2:**
\[ P_2 = \frac{V^2}{R_2} = \frac{V^2 A}{1.44\rho L} = \frac{V^2 A}{\rho L} \cdot \frac{1}{1.44} = \frac{P_1}{1.44} \]

3. **Power for Wire 3:**
\[ P_3 = \frac{V^2}{R_3} = \frac{V^2 A}{0.81\rho L} = \frac{V^2 A}{\rho L} \cdot \frac{1}{0.81} = \frac{P_1}{0.81} \]

4. **Power for Wire 4:**
\[ P_4 = \frac{V^2}{R_4} = \frac{V^2 A}{1.5\rho L} = \frac{V^2 A}{\rho L} \cdot \frac{1}{1.5} = \frac{P_1}{1.5} \]

### Step 3: Compare the power dissipated in each wire

Now we can express the power dissipated in terms of \( P_1 \):

- \( P_1 = P_1 \)
- \( P_2 = \frac{P_1}{1.44} \)
- \( P_3 = \frac{P_1}{0.81} \)
- \( P_4 = \frac{P_1}{1.5} \)

### Step 4: Rank the wires according to the power dissipated

To rank the wires from greatest to least power dissipated, we need to compare the fractions:

1. **Wire 3:** \( P_3 \) (greatest)
2. **Wire 1:** \( P_1 \)
3. **Wire 4:** \( P_4 \)
4. **Wire 2:** \( P_2 \) (least)

Thus, the order from greatest to least power dissipated is:

**Wire 3 > Wire 1 > Wire 4 > Wire 2**