Sure! Let's solve this problem step-by-step:
### Given:
- Height of mercury column, [tex]\( h = 760 \)[/tex] mm
- Density of mercury, [tex]\( \rho = 13.6 \)[/tex] gm/cm³
- Acceleration due to gravity, [tex]\( g = 9.8 \)[/tex] m/s²
### Step 1: Convert the height from mm to meters
Since 1 mm is equal to 0.001 meters:
[tex]\[ h = 760 \, \text{mm} \times 0.001 \, \frac{\text{meters}}{\text{mm}} = 0.76 \, \text{m} \][/tex]
### Step 2: Convert the density from gm/cm³ to kg/m³
Since 1 gm/cm³ is equal to 1000 kg/m³:
[tex]\[ \rho = 13.6 \, \frac{\text{gm}}{\text{cm}^3} \times 1000 \, \frac{\text{kg}}{\text{m}^3 \, \frac{\text{gm}}{\text{cm}^3}} = 13600 \, \frac{\text{kg}}{\text{m}^3} \][/tex]
### Step 3: Calculate the pressure using the formula [tex]\( P = h \rho g \)[/tex]
[tex]\[ P = h \times \rho \times g \][/tex]
[tex]\[ P = 0.76 \, \text{m} \times 13600 \, \frac{\text{kg}}{\text{m}^3} \times 9.8 \, \frac{\text{m}}{\text{s}^2} \][/tex]
### Step 4: Compute the final value:
[tex]\[ P = 0.76 \times 13600 \times 9.8 \][/tex]
[tex]\[ P = 101292.8 \, \text{Pa} \][/tex]
Therefore, the pressure exerted on the mercury column is [tex]\( 101292.8 \)[/tex] Pascals, which can be approximated as [tex]\( 1.013 \times 10^5 \)[/tex] Pascals.
