Animated Solution for Physics - Properties of Solids and Liquids: A cylinder of height $20\text{ m}$ is completely filled with water. The velocity of efflux of water (in ${\text{ms}}^{-1}$) through a small hole on the side wall of the cylinder near its bottom, is

Visualizing the Problem Imagine standing next to a massive, $20$-meter tall cylindrical water tank. It is filled to the brim. Suddenly, a tiny puncture appears near the very bottom of the tank. Water immediately begins to violently gush out. The question we are faced with is: exactly how fast is that water escaping?

This isn't just a textbook problem; it's a classic scenario that engineers and physicists deal with when designing dams, water towers, and even rocket fuel tanks. To solve it, we need to tap into one of the most elegant principles in fluid dynamics: Bernoulli's Theorem.

Bernoulli's Principle Bernoulli's theorem is essentially the law of conservation of energy applied to a flowing fluid. It tells us that for an ideal, incompressible fluid, the total mechanical energy per unit volume remains constant along any streamline.

Mathematically, it is expressed as:

Here, $P$ is the pressure energy, $\rho gh$ is the gravitational potential energy, and $\frac{1}{2}\rho v^2$ is the kinetic energy per unit volume.

To use this powerful tool, we need to choose two strategic points along the fluid's path. - Point 1: The top surface of the water in the cylinder. - Point 2: The fluid just as it exits the tiny hole at the bottom.

Let's carefully analyze the physical conditions at these two points.

At Point 1 (the top surface): - The tank is open to the atmosphere, so the pressure $P_1$ is simply the atmospheric pressure, $p_0$. - The height $h_1$ is the full height of the water column, which is $h=20\text{ m}$. - Because the hole at the bottom is described as "small," the top surface of the water will move downwards at an imperceptibly slow rate. For all practical purposes, we can assume its velocity $v_1\approx 0$.

At Point 2 (just outside the hole): - The water stream is now exposed to the open air, so the pressure $P_2$ is also the atmospheric pressure, $p_0$. - We can set this bottom level as our reference height, meaning $h_2=0$. - The velocity $v_2$ is the unknown efflux velocity, $v$, that we are trying to find.

Now, we substitute these conditions into Bernoulli's equation:

Plugging in our specific values:

Notice how beautifully the equation simplifies! The atmospheric pressure $p_0$ is present on both sides, so it cancels out. This tells us that the ambient air pressure doesn't drive the flow; gravity does.

Next, the density of the water, $\rho$, also cancels out. This is a profound realization: the speed of the exiting fluid is completely independent of its density! Whether the tank is filled with water, liquid mercury, or light oil, the efflux velocity would be exactly the same.

Rearranging the remaining terms to solve for $v$, we get:

This elegant result is famously known as Torricelli's Law. It reveals that the speed of fluid shooting out of a hole at depth $h$ is exactly the same as the speed a solid object would acquire if it were dropped in free fall from that same height $h$.

We know the height $h=20\text{ m}$ and the acceleration due to gravity $g=10{\text{ m/s}}^2$.

Substitute these into Torricelli's formula:

The water shoots out of the hole at a blistering speed of $20\text{ m/s}$!

But what if we changed the rules? If the top of the cylinder were sealed airtight, the pressure $P_1$ would rapidly drop below atmospheric pressure as water exited, eventually creating a vacuum that would stop the flow entirely. If the hole were much larger, we could no longer assume $v_1=0$, and we would need to use the Equation of Continuity ($A_1{v}_1=A_2{v}_2$) to find the exact relationship between the velocities. Always keep questioning the constraints!