Question

A stationary circular pipe of radius $R = 0.5$ m is half filled with water (density = 1000 kg/m$^3$), whereas the upper half is filled with air at atmospheric pressure, as shown in the figure. Acceleration due to gravity is $g = 9.81$ m/s$^2$. The magnitude of the force per unit length (in kN/m, rounded off to one decimal place) applied by water on the pipe section AB is $________________$.

Hint:

For hydrostatic thrust on curved surfaces, resolve the force into horizontal and vertical components using projected area and centroid depth.

Answer 1

Correct Answer: 2.2

Step 1: Understand the setup.
- Pipe radius $R = 0.5$ m
- Depth of water = $R = 0.5$ m (half filled)
- We need hydrostatic force per unit length on vertical section AB.
Step 2: Hydrostatic pressure distribution.
At depth $h$ below the free surface: \[ p(h) = \rho g h \] Total force on vertical projection AB = Hydrostatic pressure $\times$ area.
Step 3: Compute force per unit length.
Length AB = $R = 0.5$ m (water depth). \[ F = \int_0^R \rho g h \, dh \] \[ = \rho g \left[\frac{h^2}{2}\right]_0^R \] \[ = \frac{\rho g R^2}{2} \]
Step 4: Substitute values.
\[ F = \frac{1000 \times 9.81 \times (0.5)^2}{2} \] \[ = 1000 \times 9.81 \times 0.125 = 1226.25 \ \text{N/m} \] \[ = 1.23 \ \text{kN/m} \] But note: section AB is a curved surface, not flat — so we need horizontal force (projection method). Effective force on semicircle of radius 0.5: \[ F = \rho g . ( \text{Area of water cross-section}) . \frac{1}{width} \] Correct derivation → Hydrostatic thrust on semicircular vertical section per unit length: \[ F = \rho g . A_c . \bar{h} \] Where $A_c$ = semicircle area $= \frac{\pi R^2}{2}$, centroid depth = $\frac{4R}{3\pi}$. \[ F = 1000 \times 9.81 \times \frac{\pi (0.5)^2}{2} \times \frac{4(0.5)}{3\pi} \] \[ = 9810 \times 0.3927 \times 0.212 \] \[ = 1928 \ \text{N/m} \approx 1.9 \ \text{kN/m} \] Final Answer: \[ \boxed{1.9 \ \text{kN/m}} \]