A hemisphere and a solid cone have a common base. The center of mass of common structure coincides with the common base. If `R` is the radius of hemisphere and `h` is the height of the cone, then `h//R` will be

Options

A. `sqrt(3)`
B. `3`
C. `(1)/(sqrt(3))`
D. `(1)/(3)`

Explanation

To solve the problem, we need to find the ratio \( \frac{h}{R} \) where \( R \) is the radius of the hemisphere and \( h \) is the height of the cone. Given that the center of mass of the combined structure (hemisphere + cone) coincides with the common base, we can use the concept of center of mass and the volumes of the two shapes.

### Step-by-step Solution:

1. **Volume of the Hemisphere**:
The volume \( V_h \) of a hemisphere is given by the formula:
\[ V_h = \frac{2}{3} \pi R^3 \]

2. **Volume of the Cone**:
The volume \( V_c \) of a cone is given by the formula:
\[ V_c = \frac{1}{3} \pi R^2 h \]

3. **Mass of the Hemisphere and Cone**:
Assuming both the hemisphere and cone have the same density \( \rho \):
- Mass of the hemisphere \( m_h = \rho V_h = \rho \left(\frac{2}{3} \pi R^3\right) \)
- Mass of the cone \( m_c = \rho V_c = \rho \left(\frac{1}{3} \pi R^2 h\right) \)

4. **Center of Mass Calculation**:
The center of mass \( y_{cm} \) of the combined structure can be calculated using the formula:
\[ y_{cm} = \frac{m_h \cdot y_h + m_c \cdot y_c}{m_h + m_c} \]
where \( y_h \) is the distance from the base to the center of mass of the hemisphere, and \( y_c \) is the distance from the base to the center of mass of the cone.

- The center of mass of the hemisphere is at a distance of \( \frac{3R}{8} \) from the flat base.
- The center of mass of the cone is at a distance of \( \frac{h}{3} \) from the base.

5. **Setting the Center of Mass to the Base**:
Given that the center of mass coincides with the base, we set \( y_{cm} = 0 \):
\[ 0 = \frac{m_h \cdot \frac{3R}{8} + m_c \cdot \frac{h}{3}}{m_h + m_c} \]
This implies:
\[ m_h \cdot \frac{3R}{8} + m_c \cdot \frac{h}{3} = 0 \]

6. **Substituting Masses**:
Substitute \( m_h \) and \( m_c \):
\[ \left(\rho \cdot \frac{2}{3} \pi R^3\right) \cdot \frac{3R}{8} + \left(\rho \cdot \frac{1}{3} \pi R^2 h\right) \cdot \frac{h}{3} = 0 \]
Simplifying this gives:
\[ \frac{2\pi R^4}{8} + \frac{\pi R^2 h^2}{9} = 0 \]

7. **Equating and Solving for \( \frac{h}{R} \)**:
Rearranging and simplifying leads to:
\[ \frac{2R^4}{8} = -\frac{R^2 h^2}{9} \]
Dividing both sides by \( R^2 \) (assuming \( R \neq 0 \)):
\[ \frac{2R^2}{8} = -\frac{h^2}{9} \]
This results in:
\[ h^2 = -\frac{2R^2 \cdot 9}{8} \]
Solving for \( \frac{h}{R} \):
\[ \frac{h}{R} = \sqrt{\frac{2 \cdot 9}{8}} = \sqrt{\frac{18}{8}} = \sqrt{\frac{9}{4}} = \frac{3}{2} \]

### Final Result:
Thus, the ratio \( \frac{h}{R} \) is:
\[ \frac{h}{R} = \frac{3}{2} \]