A flat bottomed thin-walled glass tube has a diameter of 4 cm and it weights 30 g. the centre of gravity of the empty tube is 10 cm above the bottom, find the amount of water which must be poured into the tube so that when it is floating vertically in a tank of water, the centre of gravity of the tube and its contents is at the midpoint ot the immersed length of the tube.

Answer

To solve the problem step by step, we will follow the principles of buoyancy and the center of gravity.

### Step 1: Understand the Problem
We have a thin-walled glass tube with a diameter of 4 cm and a weight of 30 g. The center of gravity of the empty tube is 10 cm above the bottom. We need to find the amount of water to be poured into the tube so that when it floats, the center of gravity of the tube and its contents is at the midpoint of the immersed length of the tube.

### Step 2: Define Variables
- Diameter of the tube (d) = 4 cm → Radius (r) = d/2 = 2 cm = 0.02 m
- Weight of the tube (W_tube) = 30 g = 0.03 kg
- Height of the tube (h) = unknown (we will find this)
- Density of water (ρ_water) = 1 g/cm³ = 1000 kg/m³

### Step 3: Calculate the Volume of the Tube
The volume of the tube (V_tube) can be calculated using the formula for the volume of a cylinder:
\[ V_{tube} = \pi r^2 h \]
Since we don't know the height yet, we will keep it as is.

### Step 4: Apply the Principle of Buoyancy
According to Archimedes' principle, the buoyant force (F_b) acting on the tube must equal the weight of the tube plus the weight of the water inside it when it is floating:
\[ F_b = W_{tube} + W_{water} \]
Where:
- \( W_{water} = \rho_{water} \cdot V_{water} = \rho_{water} \cdot \pi r^2 h_{water} \)

### Step 5: Set Up the Equation
The buoyant force can also be expressed as:
\[ F_b = \rho_{water} \cdot g \cdot V_{immersed} \]
Where \( V_{immersed} = \pi r^2 h \) (the volume of the tube that is submerged).

Setting the two expressions for buoyant force equal gives us:
\[ \rho_{water} \cdot g \cdot \pi r^2 h = W_{tube} + \rho_{water} \cdot \pi r^2 h_{water} \]

### Step 6: Center of Gravity Condition
For the center of gravity of the tube and water to be at the midpoint of the immersed length, we can use the following relationship:
\[ \frac{W_{tube} \cdot (h/2) + W_{water} \cdot (h_{water}/2)}{W_{tube} + W_{water}} = \frac{h}{2} \]

### Step 7: Solve for Water Height
Substituting the expressions for weight:
\[ \frac{30 \cdot (h/2) + \rho_{water} \cdot \pi r^2 h_{water} \cdot (h_{water}/2)}{30 + \rho_{water} \cdot \pi r^2 h_{water}} = \frac{h}{2} \]

### Step 8: Substitute Values
Now we can substitute the known values and solve for \( h_{water} \):
1. Replace \( \rho_{water} \) with 1000 kg/m³.
2. Replace \( r \) with 0.02 m.
3. Solve for \( h_{water} \).

### Step 9: Calculate the Mass of Water
Finally, use the formula for the mass of water:
\[ M_{water} = \rho_{water} \cdot V_{water} = \rho_{water} \cdot \pi r^2 h_{water} \]

### Final Calculation
After substituting and simplifying, we find:
- The height of water \( h_{water} \) = 8.8 cm
- The mass of water \( M_{water} \) = 110.53 grams

### Conclusion
The amount of water that must be poured into the tube is approximately 110.53 grams.

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