Q.The cylindrical tube of a spray pump has a cross-section of $8.0\ \text{cm}^{2}$ one end of which has $40$ fine holes each of diameter $1.0\ \text{mm}$. If the liquid flow inside the tube is $1.5\ \text{m min}^{-1}$, what is the speed of ejection of the liquid through the holes?

Concept and Intuition

For an incompressible fluid, the total volume flow rate into any junction must equal the total volume flow rate out of it (conservation of mass, equation of continuity: $\sum A_{in}v_{in} = \sum A_{out}v_{out}$). Here, the single incoming stem of area $A$ with speed $v_1$ splits into two outgoing branches — one of area $A$ with speed $v_2$, and one of area $1.5A$ with the unknown speed $v$.

Step-by-Step Solution

1. Write the continuity equation: $Av_1 = Av_2 + (1.5A)v$.
2. Substitute known values ($v_1=3\,\mathrm{m/s}$, $v_2=1.5\,\mathrm{m/s}$): $A(3) = A(1.5) + 1.5A\,v$. …

Concept and Intuition

For an incompressible fluid in steady flow, the volume flow rate must be the same everywhere along the flow path. Since the flow splits into 50 identical holes at the exit, the tube's flow rate must equal the sum of the flow rates through all 50 holes.

Step-by-Step Solution

1. Tube radius $R=2$ cm $=0.02$ m, so $A_{tube}=\pi R^2=\pi(0.02)^2=4\pi\times10^{-4}\ \text{m}^2$.
2. Each hole radius $r=0.4$ mm $=4\times10^{-4}$ m, so $A_{hole}=\pi r^2=\pi(4\times10^{-4})^2=16\pi\times10^{-8}\ \text{m}^2$.
3. Total hole area: $50A_{hole}=800\pi\times10^{-8}=8\pi\times10^{-6}\ \text{m}^2$. …

Concept and Intuition

As a tank drains through a small hole, the exit speed (and hence the rate of fall of the level) is proportional to $\sqrt{h}$ (Torricelli's law), so the level falls slower as $h$ decreases. This means it takes progressively less time to drain each successive equal-height layer of water — quite counter to naive intuition that a slower-emptying tank should take more time near the bottom, but the volume being drained also shrinks.

Step-by-Step Solution

1. Torricelli's law gives $\dfrac{dh}{dt} = -k\sqrt{h}$ for a constant $k$ depending on the hole/tank geometry.
2. Separating variables and integrating from $h_1$ to $h_2$: $\displaystyle\int_{h_1}^{h_2}\!\frac{dh}{\sqrt h}=-k\!\int_0^{t}\!dt \Rightarrow t=\frac{2}{k}\left(\sqrt{h_1}-\sqrt{h_2}\right)$.
3. Time to fall from $h$ to $h/2$: $t_1 = \dfrac{2}{k}\left(\sqrt h - \sqrt{h/2}\right)$. …

Concept and Intuition

For a fluid flowing steadily through a pipe of changing cross-section, mass conservation demands the volume flow rate stays constant: $A_1v_1=A_2v_2$ (the equation of continuity). A narrower pipe (smaller area) must therefore carry a faster stream.

Step-by-Step Solution

1. Area $A=\dfrac{\pi d^2}{4}$, so $A\propto d^2$; the equation of continuity becomes $d_1^2v_1=d_2^2v_2$.
2. $v_2=v_1\left(\dfrac{d_1}{d_2}\right)^2=2\left(\dfrac{5}{3}\right)^2$. …

Concept and Intuition

Water draining from a hole at depth $h$ leaves with speed $v=\sqrt{2gh}$ (Torricelli), the same as free fall from height $h$. The volume flow rate is the exit speed times the hole's cross-sectional area, $Q=Av$.

Step-by-Step Solution

1. Efflux speed: $v=\sqrt{2gh}=\sqrt{2\times10\times10}=\sqrt{200}=14.14\ \mathrm{m\,s^{-1}}$.
2. Hole radius $r=\dfrac{4\ \mathrm{mm}}{2}=2\times10^{-3}\ \mathrm{m}$; area $A=\pi r^2=\pi(2\times10^{-3})^2=1.256\times10^{-5}\ \mathrm{m^2}$.
3. Flow rate: $Q=Av=1.256\times10^{-5}\times14.14\approx1.77\times10^{-4}\ \mathrm{m^3s^{-1}}$.
4. The matching printed mantissa is $1.77$, i.e. option (C).

Common Mistakes

• Using the diameter instead of the radius in the area, giving four times too large an area. …

Concept and Intuition

For an incompressible fluid flowing through a tube of varying cross-section, mass conservation requires the volume flow rate to be the same everywhere: $A_1v_1=A_2v_2$. A narrower tube forces the fluid to speed up. Since $A=\pi d^2/4$, this becomes a relation between diameters and speeds.

Step-by-Step Solution

1. Continuity: $A_1v_1=A_2v_2 \Rightarrow d_1^2v_1=d_2^2v_2$.
2. Solve for $d_2$: $d_2=d_1\sqrt{\dfrac{v_1}{v_2}}$. …

Concept and Intuition

For flow through a pipe, the volumetric flow rate is $Q = Av$, where $A$ is the pipe's cross-sectional area and $v$ is the (average) flow speed. Multiplying by elapsed time gives total volume collected.

Step-by-Step Solution

1. Radius of pipe: $r = 10$ cm $= 0.1$ m; area $A = \pi r^2 = \pi(0.1)^2 = 0.01\pi$ m².
2. Speed: $v = \dfrac{50}{\pi}$ cm/s $= \dfrac{0.5}{\pi}$ m/s.
3. Flow rate: $Q = Av = 0.01\pi\times\dfrac{0.5}{\pi} = 0.005$ m³/s (the $\pi$ cancels neatly). …

Concept and Intuition

For an incompressible fluid in steady flow, the volume flow rate is conserved: $A_1v_1=A_2v_2$ (equation of continuity). Here the "inlet" is the tube cross-section and the "outlet" is the combined area of all 30 tiny holes.

Step-by-Step Solution

1. Tube speed: $v_1=1.2\ m/min=\dfrac{1.2}{60}=0.02\ m.s^{-1}$; tube area $A_1=6\ cm^2$.
2. Each hole diameter $=1\ mm=0.1\ cm$, so hole area $=\dfrac{\pi}{4}(0.1)^2=0.0025\pi\ cm^2$.
3. Total hole area for 30 holes: $A_2=30\times0.0025\pi=0.075\pi\ cm^2$.
4. Continuity: $v_2=\dfrac{A_1v_1}{A_2}=\dfrac{6\times0.02}{0.075\pi}=\dfrac{0.12}{0.075\pi}=\dfrac{1.6}{\pi}\ m.s^{-1}=\dfrac{8}{5\pi}\ m.s^{-1}$. …

Concept and Intuition

Water keeps rising in the tank as long as inflow exceeds outflow. The outflow rate through the hole increases with the height of water above it (via Torricelli's theorem, $v=\sqrt{2gh}$), so as the tank fills, the outflow rate grows. The maximum (steady-state) height is reached exactly when the outflow rate catches up to and equals the constant inflow rate — beyond that point the water level stops rising.

Step-by-Step Solution

1. Efflux velocity from the hole at water height $h$: $v=\sqrt{2gh}$ (Torricelli's law).
2. Outflow rate $=a v = a\sqrt{2gh}$, where $a=1\,cm^2$. …

Concept and Intuition

For an idealized tank (infinite cross-section), Torricelli's law gives $v^2=2gh$. But here the ratio of hole area to tank area is a given, non-negligible $0.1$, so we must use the equation-of-continuity-corrected version, which accounts for the fact that the free surface of the tank is also moving down as water exits.

Step-by-Step Solution

1. Height of water above the hole: $h = 3\ m - 0.525\ m = 2.475\ m$.
2. Using the corrected Torricelli formula: $v^2 = \dfrac{2gh}{1-\left(\dfrac{a}{A}\right)^2}$, where $\dfrac{a}{A}=0.1$.
3. Numerator: $2gh = 2\times10\times2.475 = 49.5$. …