Find the surface energy of water kept in a cylindrical vessel of radius 6.0 cm. Surface tension of water `=0.075Jm^-2`.

Explanation

To find the surface energy of water kept in a cylindrical vessel, we can follow these steps:

### Step 1: Understand the formula for surface energy
The surface energy (U) of a liquid can be calculated using the formula:
\[ U = \text{Surface Tension} \times \text{Area} \]

### Step 2: Identify the area of the surface
For a cylindrical vessel, the surface area of the liquid in contact with air is the area of the circular top surface. The area (A) of a circle is given by:
\[ A = \pi r^2 \]
where \( r \) is the radius of the circle.

### Step 3: Convert the radius into meters
The radius given in the problem is 6.0 cm. We need to convert this into meters:
\[ r = 6.0 \, \text{cm} = 6.0 \times 10^{-2} \, \text{m} \]

### Step 4: Calculate the area
Now, we can calculate the area using the radius:
\[ A = \pi (6.0 \times 10^{-2})^2 \]
\[ A = \pi (3.6 \times 10^{-3}) \]
\[ A \approx 3.6 \times 10^{-3} \pi \, \text{m}^2 \]

### Step 5: Substitute the values into the surface energy formula
Now we can substitute the surface tension and the area into the surface energy formula. The surface tension of water is given as \( 0.075 \, \text{J/m}^2 \):
\[ U = 0.075 \, \text{J/m}^2 \times (3.6 \times 10^{-3} \pi) \]

### Step 6: Calculate the surface energy
Calculating the above expression:
\[ U \approx 0.075 \times 3.6 \times 10^{-3} \times \pi \]
\[ U \approx 0.075 \times 3.6 \times 10^{-3} \times 3.14 \]
\[ U \approx 0.075 \times 1.13184 \times 10^{-2} \]
\[ U \approx 8.4 \times 10^{-4} \, \text{J} \]

### Final Answer
The surface energy of water kept in the cylindrical vessel is approximately:
\[ U \approx 8.4 \times 10^{-4} \, \text{J} \]

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