A hemispherical portion of a radius R is removed from the bottom of a cylinder of radius R. The volume of the remaining cylinder is V and its mass is 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 of h below the liquid surface. THe force on the bottom of the cylinder by the liquid is

Options

A. M g
B. `Mg - V rho` g
C. `Mg + Pi R^(2)hg rho`
D. `rho g (V + Pi R^(2) h)`

Explanation

To solve the problem, we will analyze the forces acting on the cylinder and derive the expression for the force on the bottom of the cylinder by the liquid.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Cylinder:**
- The force acting on the bottom of the cylinder due to the liquid, denoted as \( F_1 \).
- The force acting downward on the top of the cylinder due to the liquid, denoted as \( F_2 \).
- The buoyant force acting upward on the cylinder, which can be expressed as \( F_b \).

2. **Apply the Principle of Equilibrium:**
- Since the cylinder is in equilibrium (it stays vertical), we can write the equation:
\[ F_1 - F_2 = F_b \]

3. **Express the Buoyant Force:**
- The buoyant force \( F_b \) can be expressed in terms of the volume of liquid displaced:
\[ F_b = V \cdot \rho \cdot g \]
where \( V \) is the volume of the displaced liquid, \( \rho \) is the density of the liquid, and \( g \) is the acceleration due to gravity.

4. **Calculate the Force \( F_2 \):**
- The force \( F_2 \) acting downward on the top of the cylinder can be calculated using the pressure at depth \( h \):
\[ F_2 = P \cdot A = (h \cdot \rho \cdot g) \cdot (\pi R^2) \]
where \( A = \pi R^2 \) is the cross-sectional area of the cylinder.

5. **Substituting \( F_2 \) into the Equilibrium Equation:**
- Substitute \( F_2 \) into the equilibrium equation:
\[ F_1 - (h \cdot \rho \cdot g \cdot \pi R^2) = V \cdot \rho \cdot g \]

6. **Rearranging to Find \( F_1 \):**
- Rearranging the equation gives:
\[ F_1 = V \cdot \rho \cdot g + h \cdot \rho \cdot g \cdot \pi R^2 \]

7. **Factor Out Common Terms:**
- Factor out \( \rho \cdot g \):
\[ F_1 = \rho \cdot g \left( V + h \cdot \pi R^2 \right) \]

### Final Answer:
The force on the bottom of the cylinder by the liquid is:
\[ F_1 = \rho \cdot g \left( V + h \cdot \pi R^2 \right) \]