A cylindrical vessel open at the top is `20cm` high and `10 cm` in diameter. A circular hole of cross-sectional area `1cm^(2)` is cut at the centre of the bottom of the vessel. Water flows from a tube above it into the vessel at the rate of `10^(2)cm^(3)//s`. The height of water in the vessel under steady state is (Take `g=10m//s^(2))`.

Options

A. `20cm`
B. `15cm`
C. `10cm`
D. `5cm`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the problem
We have a cylindrical vessel that is 20 cm high and 10 cm in diameter. There is a hole at the bottom with a cross-sectional area of 1 cm², and water is entering the vessel at a rate of 100 cm³/s. We need to find the steady-state height of the water in the vessel.

### Step 2: Identify the flow rates
Let:
- \( Q_{in} \) = flow rate of water entering the vessel = 100 cm³/s
- \( A_{orifice} \) = cross-sectional area of the hole = 1 cm²
- \( h \) = height of water in the vessel (in cm)
- \( g \) = acceleration due to gravity = 10 m/s² = 1000 cm/s² (since we need to convert to CGS units)

### Step 3: Calculate the velocity of water exiting the hole
Using Torricelli's theorem, the velocity \( v \) of water exiting the hole can be given by:
\[ v = \sqrt{2gh} \]
Substituting \( g \) in CGS units:
\[ v = \sqrt{2 \times 1000 \times h} = \sqrt{2000h} \]

### Step 4: Calculate the flow rate of water exiting the hole
The flow rate \( Q_{out} \) through the hole can be expressed as:
\[ Q_{out} = A_{orifice} \cdot v \]
Substituting the values:
\[ Q_{out} = 1 \cdot \sqrt{2000h} = \sqrt{2000h} \]

### Step 5: Set up the equation for steady state
At steady state, the flow rate entering the vessel equals the flow rate exiting the vessel:
\[ Q_{in} = Q_{out} \]
Substituting the known values:
\[ 100 = \sqrt{2000h} \]

### Step 6: Square both sides to eliminate the square root
\[ 100^2 = 2000h \]
\[ 10000 = 2000h \]

### Step 7: Solve for \( h \)
\[ h = \frac{10000}{2000} = 5 \text{ cm} \]

### Conclusion
The height of water in the vessel under steady state is **5 cm**.

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