---
title: "Pressure on circular plate (2.5 m)"
subject: "Hydro Power"
chapter: 1
key: "semester_6/hydro_power:circular-plate-2-5m"
source: https://www.bctengineer.com/q/semester_6/hydro_power/circular-plate-2-5m
site: BCT Engineering
---

# Pressure on circular plate (2.5 m)

Hydro Power · Chapter 1: Introduction to Basic Fluid Mechanics and Hydraulics

- Asked in: 2080 Chaitra, 6 marks
- Video: [Problem on Vertical Circular Plate Immersed in Water - Fluid Mechanics 1](https://www.youtube.com/watch?v=vJcFaNG9QPA) — Ekeeda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/circular-plate-2-5m
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-1

## Question

A circular plate of 2.5 m diameter is immersed in liquid of specific gravity 0.9, it's greatest and least depth below the free surface being 3 m and 1 m respectively. Find the total pressure on one face of the plate and position of center of pressure.

## Short answer

Inclined circular plate: sinθ = 0.8, centroid depth 2 m. Total pressure F = ρg A h̅ = 86.68 kN. Centre of pressure depth h* = h̅ + (I_G sin²θ)/(A h̅) = 2.125 m.

## Answer

The plate is not vertical because its depth range (3 − 1 = 2 m) is less than its diameter, so it is inclined. Use the formulas for an inclined plane surface.

**Given:** $d = 2.5$ m, $S = 0.9$ so $\rho = 900$ kg/m³, depths of top and bottom points $= 1$ m and $3$ m.

```text
  free surface =====================
         :  1 m
         :     * top point
         :    / \
         :   /   \   plate, d = 2.5 m
         :  /  G  \   inclined at theta
         :   \   /
         :    \ /
         :     * bottom point (3 m)
```

#### Inclination and centroid depth
$$
\begin{aligned}
\sin\theta &= \frac{3 - 1}{2.5} = 0.8000 \\
\bar h &= \frac{1 + 3}{2} = 2\ \text{m}
\end{aligned}
$$

#### Area and second moment of area
$$
\begin{aligned}
A &= \frac{\pi}{4}(2.5)^2 = 4.9087\ \text{m}^2 \\
I_G &= \frac{\pi d^4}{64} = \frac{\pi (2.5)^4}{64} = 1.9175\ \text{m}^4
\end{aligned}
$$

#### Total pressure
$$
\begin{aligned}
F &= \rho g A \bar h = 900 \times 9.81 \times 4.9087 \times 2 \\
&= 86\,679\ \text{N} = 86.68\ \text{kN}
\end{aligned}
$$

#### Centre of pressure
$$
\begin{aligned}
h^* &= \bar h + \frac{I_G \sin^2\theta}{A \bar h} \\
&= 2 + \frac{1.9175 \times 0.6400}{4.9087 \times 2} \\
&= 2 + 0.1250 = 2.125\ \text{m}
\end{aligned}
$$

**Answer:** total pressure $F \approx 86.68$ kN; centre of pressure at a depth of $2.125$ m below the free surface (on the plate's vertical diameter, below the centroid).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The circular plate has diameter 2.5 m but its depth range is only 3 - 1 = 2 m. What does this show?
   A. The liquid is not at rest
   B. The plate is inclined to the free surface
   C. The plate is vertical
   D. The plate is horizontal
   Answer: B. If the plate were vertical its depth range would equal its diameter. A smaller range means it is tilted, with sin(theta) = 2/2.5 = 0.8. (Key point: Why the plate is inclined)

2. What is the depth of the centroid of the plate below the free surface?
   A. 2 m
   B. 2.5 m
   C. 3 m
   D. 1 m
   Answer: A. The centroid is midway between the top and bottom points: (1 + 3)/2 = 2 m. (Key point: Centroid depth)

3. Which expression gives the total pressure on the plate (rho = 900 kg/m^3)?
   A. rho g A x (centroid depth)
   B. rho g A x (greatest depth)
   C. rho A x (centroid depth)
   D. rho g I_G x (centroid depth)
   Answer: A. F = rho g A h_bar with h_bar the centroid depth: 900 x 9.81 x 4.9087 x 2 = 86.68 kN. (Key point: Total pressure formula)

4. Which formula gives the depth of the centre of pressure for this inclined plate?
   A. h* = I_G sin(theta) / (A h_bar)
   B. h* = h_bar - I_G sin^2(theta) / (A h_bar)
   C. h* = h_bar + I_G / (A h_bar)
   D. h* = h_bar + I_G sin^2(theta) / (A h_bar)
   Answer: D. For an inclined plane the second term carries sin^2(theta): 2 + 1.9175 x 0.64 / (4.9087 x 2) = 2.125 m, below the centroid. (Key point: Centre of pressure for inclined plane)

## Close variants set in other exams

- [Pressure on circular plate (2.6 m)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Acircular-plate-2-6m) — Hydro Power, chapter 1, asked 2×

## Related in Hydro Power

- [Pressure on circular plate (2.2 m)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Acircular-plate-2-2m) — Hydro Power, chapter 1
- [Centre of pressure below centroid proof](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Acentre-pressure-below-centroid) — Hydro Power, chapter 1
- [Pressure on isosceles triangular plate](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Atriangular-plate-pressure) — Hydro Power, chapter 1
- [Pressure on vertical rectangular plate 2 m × 3 m](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Arectangular-plate-2x3) — Hydro Power, chapter 1
- [Pressure on inclined rectangular plate](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Ainclined-rectangular-plate) — Hydro Power, chapter 1
- [Pressure on rectangular plate 3 m × 5 m](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Arectangular-plate-3x5) — Hydro Power, chapter 1

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