A vessel in the shape of a hollow hemisphere surmounted by a once is held with the axis vertical and vertex uppermost. If it be filled with a liquid so as to submerge half the axis of the cone in the liquid, and the height of the cone be double the radius of its base, find the liquid on the vessel is x times the weight of the liquid that the hemisphere can hold.

Options

A. `15//8`
B. `1//8`
C. `5//8`
D. `15//2`

Explanation

To solve the problem, we need to find the ratio \( x \) of the weight of the liquid in the vessel to the weight of the liquid that the hemisphere can hold. We will break down the problem step by step:

### Step 1: Understand the Geometry
The vessel consists of a hollow hemisphere and a cone on top. Let the radius of the hemisphere be \( r \). The height of the cone is given as twice the radius of its base, which is also \( r \). Therefore, the height of the cone is \( 2r \).

### Step 2: Determine the Volume of Liquid in the Vessel
The liquid fills the cone to a height of \( r \) (half the height of the cone). We need to find the volume of the liquid in both the cone and the hemisphere.

**Volume of the cone filled with liquid:**
The volume \( V \) of a cone is given by:
\[ V = \frac{1}{3} \pi r^2 h \]
For our cone, the radius at the height \( r \) can be determined using similar triangles. The radius at height \( r \) is \( \frac{r}{2} \) (since the height of the cone is \( 2r \)). Thus, the volume of the liquid in the cone is:
\[ V_{\text{cone}} = \frac{1}{3} \pi \left(\frac{r}{2}\right)^2 r = \frac{1}{3} \pi \frac{r^2}{4} r = \frac{\pi r^3}{12} \]

### Step 3: Volume of the Hemisphere
The volume of the hemisphere is given by:
\[ V_{\text{hemisphere}} = \frac{2}{3} \pi r^3 \]

### Step 4: Total Volume of Liquid in the Vessel
The total volume of liquid in the vessel is the sum of the volume of the liquid in the cone and the volume of the hemisphere:
\[ V_{\text{total}} = V_{\text{cone}} + V_{\text{hemisphere}} = \frac{\pi r^3}{12} + \frac{2}{3} \pi r^3 \]

To add these volumes, we need a common denominator:
\[ \frac{2}{3} \pi r^3 = \frac{8}{12} \pi r^3 \]
Thus,
\[ V_{\text{total}} = \frac{\pi r^3}{12} + \frac{8 \pi r^3}{12} = \frac{9 \pi r^3}{12} = \frac{3 \pi r^3}{4} \]

### Step 5: Weight of the Liquid in the Vessel
The weight of the liquid in the vessel can be calculated using the formula:
\[ W_{\text{liquid}} = \text{density} \times \text{volume} \times g \]
Let the density of the liquid be \( \rho \). Therefore,
\[ W_{\text{liquid}} = \rho \cdot \frac{3 \pi r^3}{4} \cdot g \]

### Step 6: Weight of the Liquid the Hemisphere Can Hold
The weight of the liquid that the hemisphere can hold is:
\[ W_{\text{hemisphere}} = \rho \cdot \frac{2}{3} \pi r^3 \cdot g \]

### Step 7: Find the Ratio \( x \)
We need to find the ratio \( x \) such that:
\[ W_{\text{liquid}} = x \cdot W_{\text{hemisphere}} \]
Substituting the expressions we derived:
\[ \rho \cdot \frac{3 \pi r^3}{4} \cdot g = x \cdot \left(\rho \cdot \frac{2}{3} \pi r^3 \cdot g\right) \]
Canceling out common terms:
\[ \frac{3}{4} = x \cdot \frac{2}{3} \]
Solving for \( x \):
\[ x = \frac{3}{4} \cdot \frac{3}{2} = \frac{9}{8} \]

### Conclusion
Thus, the value of \( x \) is:
\[ \boxed{\frac{9}{8}} \]