The height of mercury column in a barometer in a Calcutta laboratory was recorded to be 75 cm. Calculate this pressure in SI and CGS units the following data, Specific gravity of mercury = 13.6, Density of `water = 10^3 kg/m^3, g=9.8 m/s^2` at Calcutta. Pressure `=h rho g` in usual symbols.

Explanation

To solve the problem, we need to calculate the pressure exerted by a mercury column of height 75 cm in both SI and CGS units. We will use the formula for pressure:

\[ P = h \cdot \rho \cdot g \]

Where:
- \( P \) = pressure
- \( h \) = height of the mercury column
- \( \rho \) = density of mercury
- \( g \) = acceleration due to gravity

### Step 1: Convert the height from cm to meters
Given:
- Height \( h = 75 \, \text{cm} \)

To convert cm to meters:
\[ h = 75 \, \text{cm} \times \frac{1 \, \text{m}}{100 \, \text{cm}} = 0.75 \, \text{m} \]

### Step 2: Calculate the density of mercury
Given:
- Specific gravity of mercury = 13.6
- Density of water \( \rho_{\text{water}} = 10^3 \, \text{kg/m}^3 \)

The density of mercury \( \rho_{\text{mercury}} \) can be calculated as:
\[ \rho_{\text{mercury}} = \text{Specific Gravity} \times \rho_{\text{water}} \]
\[ \rho_{\text{mercury}} = 13.6 \times 10^3 \, \text{kg/m}^3 = 13600 \, \text{kg/m}^3 \]

### Step 3: Calculate the pressure in SI units
Using the formula for pressure:
\[ P = h \cdot \rho \cdot g \]
Substituting the values:
- \( h = 0.75 \, \text{m} \)
- \( \rho = 13600 \, \text{kg/m}^3 \)
- \( g = 9.8 \, \text{m/s}^2 \)

Calculating:
\[ P = 0.75 \, \text{m} \times 13600 \, \text{kg/m}^3 \times 9.8 \, \text{m/s}^2 \]
\[ P = 0.75 \times 13600 \times 9.8 \]
\[ P = 10010 \, \text{N/m}^2 \]
\[ P \approx 10 \times 10^4 \, \text{N/m}^2 \]

### Step 4: Convert pressure to CGS units
In CGS units:
- 1 N = \( 10^5 \) dyne
- 1 m = \( 100 \) cm

Thus, to convert pressure from SI to CGS:
\[ P_{\text{CGS}} = P_{\text{SI}} \times \frac{10^5 \, \text{dyne}}{1 \, \text{N}} \times \frac{1}{(100 \, \text{cm})^2} \]
\[ P_{\text{CGS}} = (10 \times 10^4) \times 10^5 \times \frac{1}{10000} \]
\[ P_{\text{CGS}} = 10 \times 10^5 \, \text{dyne/cm}^2 \]

### Final Answers:
- Pressure in SI units: \( 10 \times 10^4 \, \text{N/m}^2 \)
- Pressure in CGS units: \( 10 \times 10^5 \, \text{dyne/cm}^2 \)