Question

Determine the atmospheric pressure at a location where the barometric reading is 700 mmHg and the gravitational acceleration is $g = 10 \text{ m/s}^2$. Assume the mercury density is $14000 \text{ kg/m}^3$.

• A. 200 kPa
• B. 100 kPa
• C. 98 kPa
• D. 101.03 kPa

Hint:

In engineering exams, watch out for approximations of constants like $g = 10 \text{ m/s}^2$ and $\rho = 14000 \text{ kg/m}^3$.
Using standard values ($g = 9.81$ and $\rho = 13600$) would yield a different value, so always use the specified data.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
This question asks for the calculation of hydrostatic pressure exerted by a fluid column (mercury) in a barometer, which corresponds to the atmospheric pressure.

Step 2: Key Formula or Approach:
The pressure exerted by a static fluid column of height $h$ is given by:
\[ P = \rho \cdot g \cdot h \]
where:
$\rho$ is the density of the fluid,
$g$ is the acceleration due to gravity, and
$h$ is the height of the fluid column.

Step 3: Detailed Explanation:

• Convert the height of the mercury column from millimeters to meters:
\[ h = 700 \text{ mmHg} = 0.7 \text{ m} \]

• The given value for gravitational acceleration is $g = 10 \text{ m/s}^2$.

• The given density of mercury is $\rho = 14000 \text{ kg/m}^3$.

• Substitute these values into the hydrostatic pressure formula:
\[ P = 14000 \text{ kg/m}^3 \times 10 \text{ m/s}^2 \times 0.7 \text{ m} \]
\[ P = 98000 \text{ Pa} \]

• Convert the pressure to kilopascals (kPa):
\[ P = \frac{98000}{1000} \text{ kPa} = 98 \text{ kPa} \]

Step 4: Final Answer:
The atmospheric pressure is $98 \text{ kPa}$.