Torricelli's barometer used mercury but pascal duplicated it using French wine of density `984 kg m^(-3)` . In that case, the height of the wine column for normal atmospheric pressure is
(Take the density of mercury `=1.36xx10^(3) kg m^(-3)`)

Options

A. 5.5 m
B. 10.5 m
C. 9.8 m
D. 15 m

Explanation

To find the height of the wine column for normal atmospheric pressure using Pascal's method, we can use the hydrostatic pressure formula, which states that the pressure exerted by a fluid column is given by:

\[ P = \rho g h \]

Where:
- \( P \) is the pressure,
- \( \rho \) is the density of the fluid,
- \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)),
- \( h \) is the height of the fluid column.

### Step-by-Step Solution:

1. **Identify the known values:**
- Density of mercury, \( \rho_{Hg} = 1.36 \times 10^3 \, \text{kg/m}^3 \)
- Density of wine, \( \rho_{wine} = 984 \, \text{kg/m}^3 \)
- Height of mercury column for normal atmospheric pressure, \( h_{Hg} = 0.76 \, \text{m} \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

2. **Calculate the pressure exerted by the mercury column:**
Using the formula for pressure:
\[ P = \rho_{Hg} g h_{Hg} \]
Substitute the known values:
\[ P = (1.36 \times 10^3 \, \text{kg/m}^3)(9.8 \, \text{m/s}^2)(0.76 \, \text{m}) \]

3. **Simplify the calculation:**
First, calculate \( \rho_{Hg} g h_{Hg} \):
\[ P = 1.36 \times 10^3 \times 9.8 \times 0.76 \]
\[ P \approx 101325 \, \text{Pa} \, (\text{which is the standard atmospheric pressure}) \]

4. **Set up the equation for the wine column:**
The pressure exerted by the wine column must equal the pressure exerted by the mercury column:
\[ P = \rho_{wine} g h_{wine} \]
Thus,
\[ 101325 = 984 \times 9.8 \times h_{wine} \]

5. **Solve for \( h_{wine} \):**
Rearranging the equation gives:
\[ h_{wine} = \frac{101325}{984 \times 9.8} \]

6. **Calculate \( h_{wine} \):**
First, calculate \( 984 \times 9.8 \):
\[ 984 \times 9.8 \approx 9672.8 \]
Now substitute back:
\[ h_{wine} = \frac{101325}{9672.8} \approx 10.5 \, \text{m} \]

### Final Answer:
The height of the wine column for normal atmospheric pressure is approximately **10.5 meters**.