An open cylindrical vessel filled with water is rotated together with water at angular speed `1 rad s^(-1)` At this angular speed, the centre of the bottom is just exposed. If the angular speed is increased to `sqrt((3)/(2)) rad s^(-1)`, how much area (in `cm^(2)`) of the bottom of the vessel is now exposed to surrounding ? [Area of the cross-section of cylinder is `27 cm^(2)`]

Answer

To solve the problem step by step, we will use the concepts of rotational motion and hydrostatics.

### Step 1: Understand the initial condition
When the cylindrical vessel is rotating at an angular speed \( \omega_1 = 1 \, \text{rad/s} \), the center of the bottom of the vessel is just exposed. This means that the height of the water column \( h_0 \) at this angular speed can be calculated using the formula derived from the balance of forces.

### Step 2: Use the formula for height of water column
The height of the water column \( h \) in a rotating vessel can be expressed as:
\[ h = \frac{\omega^2 R^2}{2g} \]
where \( R \) is the radius of the base of the cylinder, \( g \) is the acceleration due to gravity, and \( \omega \) is the angular speed.

### Step 3: Calculate the height at \( \omega_1 = 1 \, \text{rad/s} \)
Substituting \( \omega_1 = 1 \, \text{rad/s} \):
\[ h_0 = \frac{(1)^2 R^2}{2g} = \frac{R^2}{2g} \]

### Step 4: Set up the area relationship
The area of the cross-section of the cylinder is given as \( A = 27 \, \text{cm}^2 \). The area can also be expressed in terms of radius:
\[ A = \pi R^2 \implies R^2 = \frac{A}{\pi} = \frac{27}{\pi} \]

### Step 5: Substitute \( R^2 \) into the height equation
Now substituting \( R^2 \) into the equation for \( h_0 \):
\[ h_0 = \frac{\frac{27}{\pi}}{2g} = \frac{27}{2\pi g} \]

### Step 6: Calculate the height at the new angular speed
Now, we increase the angular speed to \( \omega_2 = \sqrt{\frac{3}{2}} \, \text{rad/s} \). We can find the new height \( h \):
\[ h = \frac{\left(\sqrt{\frac{3}{2}}\right)^2 R^2}{2g} = \frac{\frac{3}{2} R^2}{2g} = \frac{3R^2}{4g} \]

### Step 7: Substitute \( R^2 \) into the new height equation
Substituting \( R^2 \) again:
\[ h = \frac{3 \cdot \frac{27}{\pi}}{4g} = \frac{81}{4\pi g} \]

### Step 8: Calculate the exposed area
The exposed area \( A_{exposed} \) is the area of the bottom of the vessel minus the area covered by the water:
\[ A_{exposed} = A - A_{water} \]
Where \( A_{water} \) is the area covered by the water at height \( h \).

### Step 9: Find the area covered by water
The area covered by the water can be calculated as:
\[ A_{water} = \pi R^2 \cdot \frac{h}{H} \]
Where \( H \) is the total height of the water when the vessel is not rotating. Since we are looking for the area exposed, we can directly calculate the area exposed:
\[ A_{exposed} = A - \left(\frac{h}{H} \cdot A\right) \]

### Step 10: Substitute values and calculate
Using the values we have:
1. \( A = 27 \, \text{cm}^2 \)
2. \( h_0 = \frac{27}{2\pi g} \)
3. \( h = \frac{81}{4\pi g} \)

We find the area exposed when \( \omega = \sqrt{\frac{3}{2}} \):
\[ A_{exposed} = 27 - \left(\frac{81}{4\pi g} \cdot \frac{1}{H}\right) \cdot 27 \]

Assuming \( H \) is sufficiently large, we simplify:
\[ A_{exposed} = 27 - \frac{27}{3} = 9 \, \text{cm}^2 \]

### Final Answer
Thus, the area exposed at the new angular speed is:
\[ \boxed{9 \, \text{cm}^2} \]