A test tube consists of a hemisphere and a cylinder of the same radius. The volume of water required to fill the whole tube is 2849/3 `cm^(3)` and 2618/3 `cm^(3)` of water is required to fill the tube to a level which is 2cm below the top of the tube. Find the radius of the tube and the length of its cylindrical part.

Answer

To solve the problem, we need to find the radius (r) of the test tube and the height (h) of its cylindrical part. The test tube consists of a hemisphere and a cylinder, both having the same radius.

### Step 1: Write the volume formulas
The total volume of the test tube (hemisphere + cylinder) can be expressed as:
\[ V = \text{Volume of Hemisphere} + \text{Volume of Cylinder} \]
The volume of a hemisphere is given by:
\[ V_{\text{hemisphere}} = \frac{2}{3} \pi r^3 \]
The volume of a cylinder is given by:
\[ V_{\text{cylinder}} = \pi r^2 h \]
Thus, the total volume is:
\[ V = \frac{2}{3} \pi r^3 + \pi r^2 h \]

### Step 2: Set up the equations from the given volumes
From the problem, we know:
1. The total volume of the test tube is \( \frac{2849}{3} \, \text{cm}^3 \).
2. The volume when filled to 2 cm below the top is \( \frac{2618}{3} \, \text{cm}^3 \).

Using the total volume:
\[ \frac{2}{3} \pi r^3 + \pi r^2 h = \frac{2849}{3} \quad \text{(1)} \]

For the second volume (when filled to 2 cm below the top):
The height of the cylinder when filled to 2 cm below the top is \( h - 2 \):
\[ \frac{2}{3} \pi r^3 + \pi r^2 (h - 2) = \frac{2618}{3} \quad \text{(2)} \]

### Step 3: Simplify the equations
From equation (1):
\[ \frac{2}{3} \pi r^3 + \pi r^2 h = \frac{2849}{3} \]

From equation (2):
\[ \frac{2}{3} \pi r^3 + \pi r^2 h - 2\pi r^2 = \frac{2618}{3} \]

### Step 4: Subtract equation (2) from equation (1)
Subtracting equation (2) from equation (1) gives:
\[ \left(\frac{2}{3} \pi r^3 + \pi r^2 h\right) - \left(\frac{2}{3} \pi r^3 + \pi r^2 h - 2\pi r^2\right) = \frac{2849}{3} - \frac{2618}{3} \]
This simplifies to:
\[ 2\pi r^2 = \frac{2849 - 2618}{3} \]
Calculating the right side:
\[ 2\pi r^2 = \frac{231}{3} \]
Thus,
\[ \pi r^2 = \frac{231}{6} \]

### Step 5: Solve for r
Now, we can find \( r^2 \):
\[ r^2 = \frac{231}{6\pi} \]
Taking the square root:
\[ r = \sqrt{\frac{231}{6\pi}} \approx 3.5 \, \text{cm} \]

### Step 6: Substitute r back to find h
Now substitute \( r = 3.5 \) cm back into equation (1) to find \( h \):
\[ \frac{2}{3} \pi (3.5)^3 + \pi (3.5)^2 h = \frac{2849}{3} \]
Calculating \( \frac{2}{3} \pi (3.5)^3 \):
\[ \frac{2}{3} \pi (42.875) \approx 89.82 \]
So,
\[ 89.82 + \pi (12.25) h = \frac{2849}{3} \]
Calculating \( \frac{2849}{3} \approx 949.67 \):
\[ \pi (12.25) h = 949.67 - 89.82 \]
\[ \pi (12.25) h \approx 859.85 \]
Thus,
\[ h \approx \frac{859.85}{\pi (12.25)} \approx 20 \, \text{cm} \]

### Final Answer
The radius of the tube is \( 3.5 \, \text{cm} \) and the height of its cylindrical part is \( 20 \, \text{cm} \).