At What velocity does water emerge from an orifice in a tank in which gauge pressure is `3 xx 10^(5) Nm^(-2)` before the flow starts ? (Take the density of water `=1000 kg m^(-3)`.)

Options

A. `24.5 m s^(-1)`
B. `14.5 m s^(-1)`
C. `34.5 m s^(-1)`
D. `44.5 m s^(-1)`

Explanation

To find the velocity at which water emerges from an orifice in a tank with a given gauge pressure, we can use the principles of fluid mechanics. Here’s a step-by-step solution:

### Step 1: Understand the given values
We are given:
- Gauge pressure, \( P = 3 \times 10^5 \, \text{N/m}^2 \)
- Density of water, \( \rho = 1000 \, \text{kg/m}^3 \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

### Step 2: Relate gauge pressure to height of water column
The gauge pressure at the bottom of the tank can be related to the height of the water column above the orifice using the formula:
\[ P = h \cdot \rho \cdot g \]
Where:
- \( h \) is the height of the water column.

### Step 3: Solve for height \( h \)
Rearranging the formula to find \( h \):
\[ h = \frac{P}{\rho \cdot g} \]
Substituting the known values:
\[ h = \frac{3 \times 10^5}{1000 \cdot 9.8} \]
Calculating \( h \):
\[ h = \frac{3 \times 10^5}{9800} \approx 30.612 \, \text{m} \]

### Step 4: Use Torricelli's theorem to find velocity
According to Torricelli's theorem, the velocity \( V \) of efflux of a fluid under the influence of gravity through an orifice is given by:
\[ V = \sqrt{2gh} \]
Substituting the value of \( h \):
\[ V = \sqrt{2 \cdot 9.8 \cdot 30.612} \]

### Step 5: Calculate the velocity \( V \)
Calculating:
\[ V = \sqrt{2 \cdot 9.8 \cdot 30.612} \approx \sqrt{600} \approx 24.495 \, \text{m/s} \]

### Final Answer
The velocity at which water emerges from the orifice is approximately \( 24.495 \, \text{m/s} \).

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