Consider a water tank as shown in the figure. It's cross-sectional area is 0.4 m$$^2$$. The tank has an opening B near the bottom whose cross-section area is 1 cm$$^2$$. A load of 24 kg is applied on the water at the top when the height of the water level is 40 cm above the bottom, the velocity of water coming out the opening B is v m s$$^{-1}$$. The value of $$v$$, to the nearest integer, is ___. [Take the value of $$g$$ to be 10 m s$$^{-2}$$]

We need to find the velocity $$v$$ of water coming out of the opening $$B$$ near the bottom of the tank.

1. Identify the Circuit Parameters

From the problem statement, we have the following parameters:

• Cross-sectional area of the tank at the top ($$A$$) = $$0.4\text{ m}^2$$
• Cross-sectional area of the opening $$B$$ at the bottom ($$a$$) = $$1\text{ cm}^2 = 1 \times 10^{-4}\text{ m}^2$$
• Mass of the load applied at the top ($$M$$) = $$24\text{ kg}$$
• Height of the water level ($$h$$) = $$40\text{ cm} = 0.4\text{ m}$$
• Acceleration due to gravity ($$g$$) = $$10\text{ m s}^{-2}$$
• Density of water ($$\rho$$) = $$1000\text{ kg m}^{-3}$$

2. Apply Bernoulli's Principle

Since the top cross-sectional area ($$A = 0.4\text{ m}^2$$) is vastly larger than the opening area ($$a = 10^{-4}\text{ m}^2$$), the velocity of the water level dropping at the top is negligibly close to zero ($$v_{\text{top}} \approx 0$$).

Let's set our reference level at the bottom opening $$B$$ ($$y = 0$$). Applying Bernoulli's equation between the top surface ($$A$$) and the bottom opening ($$B$$):

$$P_{\text{top}} + \rho g h + \frac{1}{2}\rho v_{\text{top}}^2 = P_{\text{bottom}} + \rho g (0) + \frac{1}{2}\rho v^2$$

Where:

• $$P_{\text{top}} = P_0 + \frac{Mg}{A}$$ (Atmospheric pressure plus the pressure due to the external load)
• $$P_{\text{bottom}} = P_0$$ (Exposed directly to atmospheric pressure)

Substituting these pressure values and setting $$v_{\text{top}} = 0$$ into the equation gives:

$$\left(P_0 + \frac{Mg}{A}\right) + \rho g h = P_0 + \frac{1}{2}\rho v^2$$

3. Solve for Velocity ($$v$$)

Cancel out the atmospheric pressure ($$P_0$$) from both sides:

$$\frac{Mg}{A} + \rho g h = \frac{1}{2}\rho v^2$$

Isolate $$v^2$$ by multiplying the entire equation by $$\frac{2}{\rho}$$:

$$v^2 = \frac{2Mg}{\rho A} + 2gh$$

Now, plug in the given numerical values:

$$v^2 = \frac{2 \times 24 \times 10}{1000 \times 0.4} + 2 \times 10 \times 0.4$$

$$v^2 = \frac{480}{400} + 8$$

$$v^2 = 1.2 + 8 = 9.2$$

Taking the square root to find $$v$$:

$$v = \sqrt{9.2} \approx 3.03\text{ m s}^{-1}$$

Conclusion

The value of $$v$$ to the nearest integer is 3.