Question

A fully filled water tank OABCD has a circular arc (AB) of radius 10 m at the bottom as shown in the following figure. The height BC is 10 m. The length OA and CD are 5 m and 15 m, respectively. The density of the water is \(\rho\) kg/m³ and the acceleration due to gravity is \(g\) m/s². The magnitude of the resultant hydrostatic force per unit width acting on AB in N/m lies between 

 

• A. \(190\rho g\) and \(200\rho g\)
• B. \(210\rho g\) and \(220\rho g\)
• C. \(230\rho g\) and \(240\rho g\)
• D. \(250\rho g\) and \(260\rho g\)

Hint:

For hydrostatic forces on curved surfaces, always resolve the force into horizontal and vertical components. The horizontal force acts on the projected vertical area, and the vertical force is the weight of the fluid directly above the surface.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The resultant hydrostatic force on a submerged curved surface is the vector sum of the horizontal and vertical components of the force.
- The horizontal component (\(F_H\)) is the hydrostatic force on the vertical projection of the curved surface.
- The vertical component (\(F_V\)) is the weight of the fluid column directly above the curved surface up to the free surface.
Step 2: Key Formula or Approach:
Resultant Force, \(F_R = \sqrt{F_H^2 + F_V^2}\)
Horizontal Force, \(F_H = \rho g \bar{h} A_{proj}\), where \(\bar{h}\) is the depth of the centroid of the projected area from the free surface, and \(A_{proj}\) is the projected vertical area.
Vertical Force, \(F_V = \rho g V\), where \(V\) is the volume of water above the curved surface.
Step 3: Detailed Explanation or Calculation:
Let's consider a unit width (\(w = 1\) m). The coordinate system origin is at O. The free surface of the water is at height \(y = 20\) m. The circular arc AB has a radius \(R = 10\) m and its center is at (5, 10).
Calculation of Horizontal Force (\(F_H\)):
The vertical projection of the arc AB is a rectangle of height 10 m (from y=0 to y=10) and width 1 m.
Projected Area, \(A_{proj} = 10 \text{ m} \times 1 \text{ m} = 10 \text{ m}^2\).
The centroid of this projected area is at a height of \(y = 10/2 = 5\) m from the bottom.
The depth of the centroid from the free surface is \(\bar{h} = 20 - 5 = 15\) m.
\[ F_H = \rho g \bar{h} A_{proj} = \rho g \times 15 \times 10 = 150 \rho g \] Calculation of Vertical Force (\(F_V\)):
The vertical force is the weight of the water column above the arc AB. This volume (per unit width) is the area enclosed by the arc AB, the vertical lines from A and B to the free surface, and the free surface itself.
This area can be seen as the area of the rectangle with corners (5,10), (15,10), (15,20), (5,20) plus the area of the quarter-circle sector A-B-E (where E is (5,10)).
Area of the rectangle above the arc = base \(\times\) height = \((15-5) \times (20-10) = 10 \times 10 = 100 \text{ m}^2\).
Area of the quarter-circle below the horizontal line through B = \( \frac{1}{4} \pi R^2 = \frac{1}{4} \pi (10)^2 = 25\pi \text{ m}^2\).
Total area above the arc AB = \(100 + 25\pi\).
Volume per unit width, \(V = (100 + 25\pi) \times 1 = 100 + 25\pi \text{ m}^3\).
\[ F_V = \rho g V = \rho g (100 + 25\pi) \] Using \(\pi \approx 3.14159\), \(F_V \approx \rho g (100 + 25 \times 3.14159) = \rho g (100 + 78.54) = 178.54 \rho g\).
Calculation of Resultant Force (\(F_R\)):
\[ F_R = \sqrt{F_H^2 + F_V^2} = \sqrt{(150 \rho g)^2 + (178.54 \rho g)^2} \] \[ F_R = \rho g \sqrt{150^2 + 178.54^2} = \rho g \sqrt{22500 + 31876.5} \] \[ F_R = \rho g \sqrt{54376.5} \approx 233.19 \rho g \] Step 4: Final Answer:
The calculated resultant hydrostatic force is approximately \(233.19 \rho g\).
Step 5: Why This is Correct:
The value \(233.19 \rho g\) lies in the range between \(230 \rho g\) and \(240 \rho g\). Therefore, option (C) is the correct answer.