A hemispherical portion of radius $R$ is removed from the bottom of a cylinder of radius $R$. The volume of the remaining cylinder is $V$ and its mass $M$. It is suspended by a string in a liquid of density $\rho$ where it stays vertical. The upper surface of the cylinder is at a depth $h$ below the liquid surface. The force on the bottom of the cylinder by the liquid is

Step 1: Identify the given parameters.

The parameters of the cylinder and liquid are:

$$\text{Cylinder radius} = R$$ $$\text{Volume of remaining cylinder} = V$$ $$\text{Liquid density} = \rho$$ $$\text{Depth of upper surface} = h$$

Step 2: Determine the net buoyant force.

Since the bottom cavity is open and filled with liquid, the volume of displaced liquid is equal to the material volume $V$.

$$F_{\text{liq,net}} = \rho g V$$

Step 3: Calculate the downward force on the top surface.

The pressure at depth $h$ acts on the top surface area $A = \pi R^2$.

$$F_{\text{top}} = \pi R^2 h \rho g$$

Step 4: Calculate the upward force on the bottom surface.

The net force exerted by the liquid is the difference between the upward force on the bottom and the downward force on the top.

$$F_{\text{liq,net}} = F_{\text{bot}} - F_{\text{top}}$$ $$F_{\text{bot}} = F_{\text{liq,net}} + F_{\text{top}}$$ $$F_{\text{bot}} = \rho g V + \pi R^2 h \rho g$$ $$F_{\text{bot}} = \rho g (V + \pi R^2 h)$$

Answer: The force on the bottom of the cylinder by the liquid is $\rho g (V + \pi R^2 h)$.