Read the following information carefully to answer the questions.
A hemispherical bowl is filled with hot water to the brim. The contents of the bowl are transferred into a cylindrical vessel whose radius is `50%` more than its height.
If diameter of the bowl is the same as that of the cylindrical vessel, then the volume of the hot water in the cylindrical vessel is.

Options

A. `60%` of the cylindrical vessel
B. `80%` of the cylindrical vessel
C. `100%` of the cylindrical vessel
D. `50%` of the cylindrical vessel

Answer

To solve the problem, we need to find the volume of hot water in the cylindrical vessel after transferring it from the hemispherical bowl. Here are the steps to arrive at the solution:

### Step 1: Understand the dimensions of the hemispherical bowl
Let the diameter of the hemispherical bowl be \( d \). Therefore, the radius \( r \) of the bowl is:
\[ r = \frac{d}{2} \]

### Step 2: Calculate the volume of the hemispherical bowl
The volume \( V \) of a hemisphere is given by the formula:
\[ V = \frac{2}{3} \pi r^3 \]
Substituting \( r = \frac{d}{2} \):
\[ V = \frac{2}{3} \pi \left(\frac{d}{2}\right)^3 = \frac{2}{3} \pi \left(\frac{d^3}{8}\right) = \frac{\pi d^3}{12} \]

### Step 3: Understand the dimensions of the cylindrical vessel
Let the height of the cylindrical vessel be \( h \). According to the problem, the radius \( R \) of the cylindrical vessel is 50% more than its height:
\[ R = h + 0.5h = 1.5h \]

### Step 4: Calculate the volume of the cylindrical vessel
The volume \( V_c \) of a cylinder is given by the formula:
\[ V_c = \pi R^2 h \]
Substituting \( R = 1.5h \):
\[ V_c = \pi (1.5h)^2 h = \pi (2.25h^2) h = 2.25 \pi h^3 \]

### Step 5: Set the volume of the bowl equal to the volume of the cylindrical vessel
Since the contents of the bowl are transferred to the cylindrical vessel, we set the volumes equal:
\[ \frac{\pi d^3}{12} = 2.25 \pi h^3 \]
Cancelling \( \pi \) from both sides:
\[ \frac{d^3}{12} = 2.25 h^3 \]
Now, substituting \( d = 2R \) (since the diameter of the bowl is the same as that of the cylindrical vessel):
\[ \frac{(2R)^3}{12} = 2.25 h^3 \]
This simplifies to:
\[ \frac{8R^3}{12} = 2.25 h^3 \implies \frac{2R^3}{3} = 2.25 h^3 \]

### Step 6: Solve for \( h \) in terms of \( R \)
Rearranging gives:
\[ h^3 = \frac{2R^3}{3 \times 2.25} = \frac{2R^3}{6.75} = \frac{8R^3}{27} \]
Taking the cube root:
\[ h = \frac{2R}{3} \]

### Step 7: Substitute \( h \) back to find the volume of the cylindrical vessel
Now substituting \( h \) back into the volume formula for the cylindrical vessel:
\[ V_c = 2.25 \pi h^3 = 2.25 \pi \left(\frac{2R}{3}\right)^3 = 2.25 \pi \left(\frac{8R^3}{27}\right) = \frac{18R^3}{27} \pi = \frac{2R^3}{3} \pi \]

### Step 8: Find the volume of hot water in the cylindrical vessel
Since the volume of hot water transferred from the bowl is equal to the volume of the bowl, we have:
\[ V = \frac{\pi d^3}{12} \]
And since \( d = 2R \):
\[ V = \frac{\pi (2R)^3}{12} = \frac{\pi (8R^3)}{12} = \frac{2\pi R^3}{3} \]

### Conclusion
The volume of hot water in the cylindrical vessel is equal to the volume of the bowl, which is \( \frac{2\pi R^3}{3} \).