Question:
Given: Density of mercury is 13,600 kg/m\(^3\) and acceleration due to gravity is 9.81 m/s\(^2\). Atmospheric pressure is 101 kPa. In a mercury U-tube manometer, the difference between the heights of the liquid in the U-tube is 1 cm. The differential pressure being measured in pascal is \(\underline{\hspace{2cm}}\) (rounded off to the nearest integer).
Given: Density of mercury is 13,600 kg/m\(^3\) and acceleration due to gravity is 9.81 m/s\(^2\). Atmospheric pressure is 101 kPa. In a mercury U-tube manometer, the difference between the heights of the liquid in the U-tube is 1 cm. The differential pressure being measured in pascal is \(\underline{\hspace{2cm}}\) (rounded off to the nearest integer).
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To calculate the differential pressure in a U-tube manometer, use the formula \( \Delta P = \rho g h \), where \( \rho \) is the density, \( g \) is the acceleration due to gravity, and \( h \) is the height difference.
Updated On: Jan 8, 2026
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Correct Answer: 1333
Solution and Explanation
The differential pressure in a U-tube manometer is given by:
\[
\Delta P = \rho g h
\]
where \( \rho \) is the density of mercury, \( g \) is the acceleration due to gravity, and \( h \) is the height difference.
Given: \[ \rho = 13600 \, \text{kg/m}^3, g = 9.81 \, \text{m/s}^2, h = 0.01 \, \text{m}. \] Substituting the values: \[ \Delta P = 13600 \times 9.81 \times 0.01 = 1333 \, \text{Pa}. \] Thus, the differential pressure is \( 1333 \, \text{Pa} \).
Given: \[ \rho = 13600 \, \text{kg/m}^3, g = 9.81 \, \text{m/s}^2, h = 0.01 \, \text{m}. \] Substituting the values: \[ \Delta P = 13600 \times 9.81 \times 0.01 = 1333 \, \text{Pa}. \] Thus, the differential pressure is \( 1333 \, \text{Pa} \).
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