The density of liquid gallium at `30^(@)C` is 6.095 g/mL. Because of its wide liquid range (30 to `2400^(@)C`), gallium is used as a barometer fluid at high temperature. What height (in cm) of gallium will be observed on a day when the mercury barometer reads 740 torr? `
` (The density of mercury is 13.6 g/mL.)

Options

A. 322
B. 285
C. 165
D. 210

Explanation

To solve the problem, we need to find the height of liquid gallium that corresponds to a pressure of 740 torr, using the given densities of gallium and mercury. We will use the formula that relates pressure, height, density, and gravitational acceleration.

### Step-by-Step Solution:

1. **Convert Pressure from Torr to Pascal:**
- Given that \(1 \text{ Torr} = 133.322 \text{ Pascal}\), we can convert 740 Torr to Pascal.
\[ P = 740 \text{ Torr} \times 133.322 \text{ Pascal/Torr} = 98650 \text{ Pascal} \]

2. **Identify the Density of Gallium:**
- The density of liquid gallium at \(30^\circ C\) is given as \(6.095 \text{ g/mL}\). We need to convert this to kg/m³ for consistency in units.
\[ \text{Density of gallium} = 6.095 \text{ g/mL} = 6095 \text{ kg/m}^3 \]

3. **Use the Pressure-Height-Density Relation:**
- The formula relating pressure (\(P\)), height (\(h\)), density (\(\rho\)), and gravitational acceleration (\(g\)) is:
\[ P = h \cdot \rho \cdot g \]
- Rearranging this formula to solve for height (\(h\)):
\[ h = \frac{P}{\rho \cdot g} \]

4. **Substitute the Known Values:**
- We know \(P = 98650 \text{ Pascal}\), \(\rho = 6095 \text{ kg/m}^3\), and \(g \approx 9.8 \text{ m/s}^2\).
\[ h = \frac{98650}{6095 \cdot 9.8} \]

5. **Calculate the Height:**
- Performing the calculation:
\[ h = \frac{98650}{59751} \approx 1.65 \text{ meters} \]

6. **Convert Height to Centimeters:**
- Since we need the height in centimeters:
\[ h = 1.65 \text{ m} \times 100 \text{ cm/m} = 165 \text{ cm} \]

### Final Answer:
The height of gallium observed when the mercury barometer reads 740 torr is **165 cm**.