There is a small hole at the bottom of tank filled with water. If total pressure at the bottom is `3 atm(1 atm=10^(5)Nm^(-2))`, then find the velocity of water flowing from hole.

Explanation

To find the velocity of water flowing from a hole at the bottom of a tank filled with water, we can use the principle of fluid dynamics, specifically Torricelli's theorem. The velocity of the fluid can be derived from the pressure difference.

### Step-by-Step Solution:

1. **Identify the Given Data**:
- Total pressure at the bottom of the tank, \( P = 3 \, \text{atm} \)
- 1 atm = \( 10^5 \, \text{N/m}^2 \)
- Density of water, \( \rho = 10^3 \, \text{kg/m}^3 \)

2. **Convert Pressure to SI Units**:
- Total pressure at the bottom in SI units:
\[ P = 3 \, \text{atm} = 3 \times 10^5 \, \text{N/m}^2 \]

3. **Calculate the Pressure at the Surface**:
- The pressure at the surface of the water is 1 atm, which is:
\[ P_{\text{surface}} = 1 \, \text{atm} = 1 \times 10^5 \, \text{N/m}^2 \]

4. **Determine the Change in Pressure**:
- The change in pressure (\( \Delta P \)) that causes the water to flow out of the hole is given by:
\[ \Delta P = P - P_{\text{surface}} = (3 \times 10^5) - (1 \times 10^5) = 2 \times 10^5 \, \text{N/m}^2 \]

5. **Use the Bernoulli's Equation to Find Velocity**:
- According to Bernoulli's principle, the velocity (\( V \)) of the fluid can be calculated using the formula:
\[ V = \sqrt{\frac{2 \Delta P}{\rho}} \]

6. **Substitute the Values**:
\[ V = \sqrt{\frac{2 \times (2 \times 10^5)}{10^3}} = \sqrt{\frac{4 \times 10^5}{10^3}} = \sqrt{4 \times 10^2} = \sqrt{400} = 20 \, \text{m/s} \]

7. **Conclusion**:
- The velocity of water flowing from the hole at the bottom of the tank is \( 20 \, \text{m/s} \).