3080 `cm^(3)` of water is required to fill a cylindrical vessel completely and 2310 `cm ^(3)` of water is required to fill it upto 5 cm below the top. Find :
wetted surface area of the vessel when it is half-filled with water.

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Understand the Volume Relationships
We know that:
- The total volume of the cylindrical vessel when completely filled is \( V_1 = 3080 \, \text{cm}^3 \).
- The volume when filled up to 5 cm below the top is \( V_2 = 2310 \, \text{cm}^3 \).

### Step 2: Calculate the Difference in Volume
The difference in volume when the height is reduced by 5 cm is:
\[ \Delta V = V_1 - V_2 = 3080 \, \text{cm}^3 - 2310 \, \text{cm}^3 = 770 \, \text{cm}^3 \]

### Step 3: Determine the Height of the Cylinder
Since the volume difference corresponds to a height difference of 5 cm, we can set up a proportion:
\[ \frac{\Delta V}{\Delta h} = \frac{770 \, \text{cm}^3}{5 \, \text{cm}} \]
This means that for every 5 cm of height, the volume changes by 770 cm³.

Now, we can find the total height \( h \) of the cylinder:
\[ \text{If } V_1 = \pi r^2 h \text{ and } V_2 = \pi r^2 (h - 5) \]
From the volume formula, we can express \( h \):
\[ h = \frac{V_1}{\pi r^2} \]

### Step 4: Find the Radius
Using the volume of the cylinder when it is completely filled:
\[ 3080 = \pi r^2 h \]
Substituting \( h \) with \( 20 \) cm (as calculated from the height):
\[ 3080 = \frac{22}{7} r^2 \cdot 20 \]
Rearranging gives:
\[ r^2 = \frac{3080 \times 7}{22 \times 20} \]
Calculating \( r^2 \):
\[ r^2 = \frac{21560}{440} = 49 \implies r = 7 \, \text{cm} \]

### Step 5: Calculate the Total Surface Area
The total surface area \( A \) of a cylinder is given by:
\[ A = 2\pi rh + 2\pi r^2 \]
Substituting the values \( r = 7 \, \text{cm} \) and \( h = 20 \, \text{cm} \):
\[ A = 2 \cdot \frac{22}{7} \cdot 7 \cdot 20 + 2 \cdot \frac{22}{7} \cdot 7^2 \]
Calculating:
\[ A = 2 \cdot 22 \cdot 20 + 2 \cdot 22 \cdot 7 = 880 + 308 = 1188 \, \text{cm}^2 \]

### Step 6: Calculate the Wetted Surface Area When Half-Filled
The wetted surface area when the cylinder is half-filled includes the curved surface area and the base area:
\[ \text{Wetted Surface Area} = \text{Curved Surface Area} + \text{Base Area} \]
The height when half-filled is \( \frac{h}{2} = 10 \, \text{cm} \):
\[ \text{Curved Surface Area} = 2\pi rh = 2 \cdot \frac{22}{7} \cdot 7 \cdot 10 = 440 \, \text{cm}^2 \]
The base area is:
\[ \text{Base Area} = \pi r^2 = \frac{22}{7} \cdot 49 = 154 \, \text{cm}^2 \]
Thus, the total wetted surface area is:
\[ \text{Wetted Surface Area} = 440 + 154 = 594 \, \text{cm}^2 \]

### Final Answer
The wetted surface area of the vessel when it is half-filled with water is \( 594 \, \text{cm}^2 \).

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