A stationary tank is cylindrical in shape with two hemispherical ends and is horizontal, as shown in the figure. \( R \) is the radius of the cylinder as well as of the hemispherical ends. The tank is half filled with an oil of density \( \rho \) and the rest of the space in the tank is occupied by air. The air pressure, inside the tank as well as outside it, is atmospheric. The acceleration due to gravity (\( g \)) acts vertically downward. The net horizontal force applied by the oil on the right hemispherical end (shown by the bold outline in the figure) is:
Show Hint
In problems involving pressure and forces on curved surfaces, consider using integration over the surface area to account for varying pressure with depth. For spherical and hemispherical surfaces, the geometry simplifies the integration.
The problem involves a stationary cylindrical tank with hemispherical ends, half-filled with oil and the rest occupied by air. The net horizontal force on the right hemispherical end of the tank is due to the pressure exerted by the oil. Since the oil is subject to the gravitational force, the pressure varies with depth.
To calculate the net horizontal force on the hemispherical end, we consider the pressure distribution across the surface of the hemispherical end. The force on a surface due to pressure is given by:
\[
F = \int P \, dA
\]
where \( P \) is the pressure at a given point on the surface, and \( dA \) is the differential area element. For a hemispherical end, the pressure at a depth \( h \) is \( P = \rho g h \), where \( h \) is the vertical distance from the top of the oil column.
By integrating the pressure over the surface of the hemisphere and using the geometry of the situation, the net horizontal force is found to be:
\[
F = \frac{2}{3} \rho g R^3
\]
Thus, the correct answer is (B).
