Animated Solution for Physics - Properties of Solids and Liquids: Water is flowing continuously from a tap having an internal diameter $8\times 10^{-3}\text{ m}$. The water velocity as it leaves the tap is $0.4{\text{ ms}}^{-1}$. The diameter of the water stream at a distance $2\times 10^{-1}\text{ m}$ below the tap is close to

Have you ever turned on a tap and watched the water flow down? If you look closely, the stream of water doesn't stay the same width; it tapers and gets narrower the further it falls. This isn't just a random occurrence—it's a beautiful, everyday demonstration of the fundamental laws of fluid mechanics!

In this problem, we are going to use two powerful principles: the Equation of Continuity and Bernoulli's Theorem (or simply the kinematics of free fall) to calculate exactly how much the stream narrows.

Imagine the water leaving the tap. We are given its initial conditions: The initial diameter is $D_1=8\times 10^{-3}\text{ m}$. The initial velocity is $v_1=0.4\text{ m/s}$.

We want to find the new diameter $D_2$ after the water has fallen a vertical distance of $h=0.2\text{ m}$. As the water falls, gravity pulls it downwards, causing it to accelerate.

Water is an incompressible fluid. This means that the volume of water passing through any horizontal cross-section of the stream per second must be exactly the same. If it weren't, water would be magically created or destroyed mid-air!

Mathematically, this is expressed as the Equation of Continuity:

Because the cross-sectional area of a circle is $A=\frac{\pi D^2}{4}$, the area is proportional to the square of the diameter. Therefore, as the velocity $v$ increases, the area $A$ (and thus the diameter $D$) must decrease to keep the product constant.

To find out how much the velocity increases, we can apply Bernoulli's Theorem between the top of the stream (just leaving the tap) and the point $0.2\text{ m}$ below. Since the stream is open to the air, the pressure is atmospheric at both points. The only energy conversion happening is gravitational potential energy turning into kinetic energy.

Notice that the density $\rho$ cancels out perfectly! This simplifies to the familiar third equation of kinematics for a body in free fall:

Now, let's merge our two concepts. From the continuity equation, we know that $v_2=v_1\frac{A_1}{A_2}$. Let's substitute this into our velocity equation:

Dividing the entire equation by $v_1^2$, we get:

Since the ratio of areas $\frac{A_1}{A_2}$ is equal to the ratio of the diameters squared ${\left(\frac{D_1}{D_2}\right)}^2$, we can rewrite this in terms of diameters:

This is our master equation. Now, we just need to carefully substitute the given values. Let's plug in $g=10{\text{ m/s}}^2$, $h=0.2\text{ m}$, and $v_1=0.4\text{ m/s}$:

Let's simplify the fraction on the right side. The numerator is $2\times 10\times 0.2=4$. The denominator is $0.4^2=0.16$.

Adding the $1$, our equation becomes beautifully simple:

To solve for $D_2$, we take the fourth root of both sides. The fourth root of $26$ is slightly larger than the fourth root of $16$ (which is $2$) and much smaller than the fourth root of $81$ (which is $3$). It evaluates to approximately $2.26$.

Looking at our options, this is closest to $3.6\times 10^{-3}\text{ m}$.

The next time you wash your hands, take a moment to appreciate the elegant physics governing the shape of the water stream!