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## Velocity of Efflux at The Bottom of A Tank

Let’s discuss a problem based on the velocity of efflux at the bottom of a tank. Try the problem yourself and read the solution here.

The Problem:

A large tank is filled with water. The total pressure at the bottom is (3.0atm). If a small hole is punched at the bottom, what is the velocity of efflux?

Solution:

A large tank is filled with water. The total pressure at the bottom is (3.0atm). A small hole is punched at the bottom.
Pressure at the bottom due to water coloumn $$(3-1)atm$$ $$=2atm$$ $$=2\times 10^5 Pa$$

The equation for pressure is $$P=h\rho g$$
Hence, $$h=\frac{P}{\rho g}$$ $$=\frac{2\times 10^5}{1000g}$$ $$=\frac{200}{g}$$
Hence, velocity $$v=\sqrt{2gh}$$ $$=\sqrt{\frac{200}{2g}}$$ $$=20m/s$$

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## Maximum Height of Water in A Tank With A Hole

Let’s discuss a problem based on the maximum height of water in a tank with a hole. Try the problem yourself first and then read the solution.

The Problem: Maximum Height of Water

A large tank is filled with water of (70 cm^3/s). A hole of cross-section (0.25cm^2) is punched at the bottom of the tank. Find the maximum height to which the tank can be filled.

Solution:

For the water level to remain stationary volume efflux= rate of filling = (x)
The velocity $$v=\sqrt{2gh}$$ where (g) is the acceleration due to gravity.
Hence,$$vA=\sqrt{2gh}A$$ $$=x$$ $$=70cm^3//s$$
The maximum height $$h=\frac{x^2}{2gA^2}$$ $$=\frac{70^2}{2\times 980\times(0.25)^2}$$ $$=40cm$$

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## Efflux Velocity of Fluid through a Small Orifice in a Tube

Let’s discuss a problem based on efflux velocity of fluid through a small orifice in a tube. Try it yourself first, then read the solution.

The Problem:

A horizontal tube of length (L), open at (A) and closed at (B), is filled with an ideal fluid. The end (B) has a small orifice. The tube is set in rotation in the horizontal plane with angular velocity (\omega) about a vertical axis passing through (A). Show that the efflux velocity of the fluid is given by $$v=\omega l\sqrt{\frac{2L}{l}-1}$$ where (l) is the length of the fluid.

Solution:

Consider a mass element (dm) of the fluid at a distance (x) from the vertical axis. The centrifugal force on (dm) is
$$Df=dm\omega^2x$$ $$=dm\frac{dv}{dt}$$ $$=dm \frac{dv}{dx}v$$
$$vdv=\omega^2 xdx$$
$$\int vdv=\omega^2 \int xdx$$ $$\frac{v^2}{2}=\frac{\omega^2}{2} \int_{L}^{L-l}$$
So,
$$v=\omega l\sqrt{\frac{2L}{l}-1}$$