This problem involves the flow of water from a tap and requires applying the principle of continuity and the equations of motion under gravity. Let's break it down step by step.

Step 1: Understand the Principle of Continuity

For an incompressible fluid like water, the flow rate remains constant. This is given by the equation of continuity:

where $A$ is the cross-sectional area and $v$ is the velocity of the fluid.

Step 2: Identify the Known Quantities

At the tap (point 1):

• Internal diameter, $d_1=8\times {10}^{-3}\text{m}$
• Radius, $r_1=\frac{d_1}{2}=4\times {10}^{-3}\text{m}$
• Area, $A_1=\pi r_1^2=\pi {(4\times {10}^{-3})}^2=16\pi \times {10}^{-6}{\text{m}}^2$
• Velocity, $v_1=0.4\text{m/s}$

At a distance $h=2\times {10}^{-1}\text{m}$ below the tap (point 2), we need to find the diameter $d_2$.

Step 3: Find the Velocity at Point 2

As water falls under gravity, its velocity increases. Using the equation of motion:

where $g=9.8\text{m/s²}$ (acceleration due to gravity).

Substitute the values:

Step 4: Apply Continuity Equation

From continuity:

So,

Step 5: Find the Diameter at Point 2

Area $A_2=\pi r_2^2$, so:

Therefore, the diameter $d_2=2r_2\approx 3.56\times {10}^{-3}\text{m}$, which is close to $3.6\times {10}^{-3}\text{m}$.

Final Answer

The diameter of the water stream at a distance $2\times {10}^{-1}\text{m}$ below the tap is close to $3.6\times {10}^{-3}\text{m}$.

Related Topics and Formulae

Equation of Continuity: For incompressible fluids, $Av=\text{constant}$. This is derived from the conservation of mass.

Equations of Motion: Under constant acceleration (like gravity), $v^2=u^2+2as$, where $u$ is initial velocity, $v$ is final velocity, $a$ is acceleration, and $s$ is displacement.

Area of Circle: $A=\pi r^2=\pi \frac{d^2}{4}$, where $d$ is diameter.