Question

Water is flowing continuously from a tap having an internal diameter $8 \times {10^{ - 3}}m$ .The water velocity as it leaves the tap is $0.4m{s^{ - 1}}$ . The diameter of the water stream at a distance $2 \times {10^{ - 1}}m$ below the tap is close to.
A. $5.0 \times {10^{ - 3}}m$
B. $7.5 \times {10^{ - 3}}m$
C. $9.6 \times {10^{ - 3}}m$
D. $3.6 \times {10^{ - 3}}m$

Answer 1

We will be using the Bernoullli’s equation for determining rate of flow in streamlined motion of liquids ; ${P_1} + \dfrac{1}{2}\rho v_1^2 + \rho g{h_1} = {P_2} + \dfrac{1}{2}\rho v_2^2 + \rho g{h_2}$ this is the bernoulli’s equation for rate of flow of liquid , after calculating velocity we will use continuity equation to determine the diameter of water stream when it is a distance below, continuity equation ; ${A_1}{v_1} = {A_2}{v_2}$ .

Complete step-by-step answer:
Now from the question
The given values are; velocity of stream as it leaves the tap ${v_1} = 0.4m{s^{ - 1}}$
Height of stream $h = 2 \times {10^{ - 1}}m$
Inner diameter of tap or diameter of water stream when it leaves the tap ${D_1} = 8 \times {10^{ - 3}}m$
Now using the Bernoulli’s equation , we have
${P_1} + \dfrac{1}{2}\rho v_1^2 + \rho g{h_1} = {P_2} + \dfrac{1}{2}\rho v_2^2 + \rho g{h_2}$
$\Rightarrow {P_0} + \dfrac{1}{2}\rho v_1^2 + \rho gh = {P_0} + \dfrac{1}{2}\rho v_2^2$
where ${P_1} = {P_2} = {P_0}$ as there is no external pressure applied hence ${P_0}$ is atmospheric pressure which is same at every place
$\rho$ is density of water and $g$ is acceleration due to gravity
and ${h_1} = h$ and ${h_2} = 0$ since it is in mid air and not specifically given
and ${v_2}$ is the velocity of stream at ${h_2} = 0$
now,
$\begin{aligned} \dfrac{1}{2}v_1^2 + gh = \dfrac{1}{2}v_2^2 \\\ \Rightarrow v_2^2 = v_1^2 + 2gh \\\ \end{aligned}$
$\Rightarrow {v_2} = \sqrt {v_1^2 + 2gh}$
Now Substituting values, we get
$\Rightarrow {v_2} = \sqrt {{{\left( {0.4} \right)}^2} + 2 \times 10 \times 2 \times {{10}^{ - 1}}}$
$\begin{gathered} \Rightarrow {v_2} = \sqrt {0.16 + 4.0} \\\ \end{gathered}$
$\Rightarrow {v_2} = \sqrt {4.16}$
$\Rightarrow {v_2} = 2.03m{s^{ - 1}}$
Now from the equation of continuity,
${A_1}{v_1} = {A_2}{v_2}$
$\dfrac{{\pi D{\kern 1pt} _1^2}}{4}{v_1} = \dfrac{{\pi D{\kern 1pt} _2^2}}{4}{v_1}$
Where ${A_1} = \dfrac{{\pi D{\kern 1pt} _1^2}}{4}$ is area of stream after leaving the tap and ${A_2} = \dfrac{{\pi D{\kern 1pt} _2^2}}{4}$ is the area of stream at ${h_2}$ and ${D_2}$ is the diameter of stream at ${h_2}$ .
hence
$D_2^2 = \dfrac{{D_1^2{v_1}}}{{{v_2}}}$
$\Rightarrow {D_2} = \sqrt {\dfrac{{{v_1}}}{{{v_2}}}} {D_1}$
$\Rightarrow D_2 = \sqrt {\dfrac{{0.4}}{{2.03}}} \left( {8 \times {{10}^{ - 3}}} \right)$
$\Rightarrow {D_2} = 0.0036m$ or ${D_2} = 3.6 \times {10^{ - 3}}m$ approx

So, the correct answer is “Option D”.

Note: Bernoulli's principle is a statement about how the speed of a fluid is related to the pressure of the fluid. Bernoulli's principle is stated as,
Bernoulli's principle : Within a horizontal flow of fluid, points of high speed of fluids will have less pressure than points of slow speed fluids.
So when the diameter of the horizontal water pipe is changing , regions where the water is moving fast will be under less pressure than regions where the water moves slow.