Question:
A vertical wall of height \(6\) m is subjected to a pressure by a liquid of height \(4\) m on one of its sides. The total pressure from the liquid surface acts at a distance of
A vertical wall of height \(6\) m is subjected to a pressure by a liquid of height \(4\) m on one of its sides. The total pressure from the liquid surface acts at a distance of
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For a vertical rectangular surface with its top at the free surface,
\[
\boxed{
h_{cp}=\frac{2h}{3}
}
\]
where \(h\) is the depth of the liquid.
Updated On: Jul 23, 2026
- \(\dfrac{4}{3}\) m
- \(2\) m
- \(\dfrac{8}{3}\) m
- \(4\) m
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Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Concept:
For a vertical rectangular surface submerged in a liquid,
the hydrostatic pressure distribution is triangular.
The centre of pressure is located at
\[
\boxed{
h_{cp}=\frac{2h}{3}
}
\]
below the free liquid surface.
where
\[
h=\text{Depth of liquid}.
\]
Step 1: Write the given data. Depth of liquid, \[ h=4\text{ m}. \]
Step 2: Calculate the centre of pressure. \[ h_{cp} = \frac{2h}{3} = \frac{2\times4}{3} = \frac{8}{3}\text{ m}. \] Thus, the resultant hydrostatic force acts \[ \boxed{\frac{8}{3}\text{ m}} \] below the liquid surface. Therefore, the correct option is \[ \boxed{(C)\;\dfrac{8}{3}\text{ m}.} \]
Step 1: Write the given data. Depth of liquid, \[ h=4\text{ m}. \]
Step 2: Calculate the centre of pressure. \[ h_{cp} = \frac{2h}{3} = \frac{2\times4}{3} = \frac{8}{3}\text{ m}. \] Thus, the resultant hydrostatic force acts \[ \boxed{\frac{8}{3}\text{ m}} \] below the liquid surface. Therefore, the correct option is \[ \boxed{(C)\;\dfrac{8}{3}\text{ m}.} \]
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