A hole is made at the bottom of the tank filled with water (density `=1000 kgm^(-3))`. If the total pressure at the bottom of the tank is three atmospheres (1 atmosphere `=10^(5) Nm^(-2))`, then the velocity of efflux is nearest to

Options

A. `sqrt(400) ms^(-1)`
B. `sqrt(200) ms^(-1)`
C. `sqrt(600) ms^(-1)`
D. `sqrt(500) ms^(-1)`

Explanation

To solve the problem of finding the velocity of efflux from a hole at the bottom of a tank filled with water, we can use Bernoulli's equation, which relates the pressure energy and kinetic energy of the fluid.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Density of water, \( \rho = 1000 \, \text{kg/m}^3 \)
- Total pressure at the bottom of the tank, \( P = 3 \, \text{atmospheres} = 3 \times 10^5 \, \text{N/m}^2 \) (since \( 1 \, \text{atmosphere} = 10^5 \, \text{N/m}^2 \))

2. **Determine the Pressure Difference:**
- The pressure at the bottom of the tank is given as 3 atmospheres. The pressure outside the hole (atmospheric pressure) is 1 atmosphere.
- Therefore, the pressure difference driving the efflux is:
\[ \Delta P = P - P_{\text{atm}} = 3P_{\text{atm}} - P_{\text{atm}} = 2P_{\text{atm}} = 2 \times 10^5 \, \text{N/m}^2 \]

3. **Apply Bernoulli’s Equation:**
- According to Bernoulli’s principle, the difference in pressure energy per unit volume is equal to the kinetic energy per unit volume:
\[ \frac{\Delta P}{\rho} = \frac{V^2}{2} \]
- Substituting the values we have:
\[ \frac{2 \times 10^5}{1000} = \frac{V^2}{2} \]

4. **Simplify the Equation:**
- Calculate the left side:
\[ \frac{2 \times 10^5}{1000} = 200 \, \text{m}^2/\text{s}^2 \]
- Thus, we have:
\[ 200 = \frac{V^2}{2} \]

5. **Solve for Velocity \( V \):**
- Multiply both sides by 2:
\[ V^2 = 400 \]
- Take the square root to find \( V \):
\[ V = \sqrt{400} = 20 \, \text{m/s} \]

### Final Answer:
The velocity of efflux is nearest to \( 20 \, \text{m/s} \).

---