Welcome to Lecture 85 of our comprehensive JEE Physics Master Series. In this lecture, we examine one of the foundational cornerstones of fluid dynamics: Bernoulli’s Principle. We explore the rigorous derivation of Bernoulli’s equation from the Work-Energy Theorem, analyze the three classic heads of fluid energy, and evaluate real-world physical and engineering applications including the Venturimeter, Torricelli’s Law of Efflux, Atomizers/Aspirators, and Aerodynamic Lift.
1. Detailed Concept Card
1.1 Assumptions for an Ideal Fluid
Bernoulli’s theorem strictly governs the motion of an ideal fluid in a steady flow regime. The physical assumptions underpinning this formulation are:
- 1. Steady (Streamline) Flow: At any fixed spatial coordinate in the fluid velocity field, the velocity, pressure, and density vectors remain independent of time: $$\frac{\partial \vec{v}}{\partial t} = 0, \quad \frac{\partial P}{\partial t} = 0, \quad \frac{\partial \rho}{\partial t} = 0$$
- 2. Incompressible: The fluid density $\rho$ is uniform and constant throughout space and time: $$\rho = \text{constant} \implies \nabla \cdot \vec{v} = 0$$
- 3. Non-Viscous (Inviscid): Internal fluid friction (viscous shear stress) between adjacent fluid layers is zero ($\eta = 0$). There is no mechanical energy dissipation into thermal internal energy.
- 4. Irrotational: Fluid elements possess zero net angular velocity about their own centers of mass: $$\vec{\omega} = \frac{1}{2}(\nabla \times \vec{v}) = 0$$ Note: For irrotational flow, the Bernoulli constant is identical across all streamlines in the entire fluid field. For rotational flow, Bernoulli’s equation is valid only along a single streamline.
1.2 Derivation from the Work-Energy Theorem
Consider a streamtube of an ideal fluid flowing between two cross-sections $1$ and $2$ of areas $A_1$ and $A_2$ at elevations $y_1$ and $y_2$ above a reference datum. In an infinitesimal time interval $\Delta t$:
- Fluid at cross-section $1$ advances by distance $\Delta x_1 = v_1 \Delta t$.
- Fluid at cross-section $2$ advances by distance $\Delta x_2 = v_2 \Delta t$.
By the Continuity Equation for an incompressible fluid:
$$\Delta V = A_1 \Delta x_1 = A_2 \Delta x_2 = A_1 v_1 \Delta t = A_2 v_2 \Delta t \implies \Delta m = \rho \Delta V$$
The net work done on the fluid element by external pressure forces is:
$$W_{\text{pressure}} = F_1 \Delta x_1 – F_2 \Delta x_2 = (P_1 A_1)(v_1 \Delta t) – (P_2 A_2)(v_2 \Delta t) = (P_1 – P_2) \Delta V$$
The work done by gravity on this mass element as it shifts from elevation $y_1$ to $y_2$ is:
$$W_{\text{gravity}} = -\Delta U_g = -\Delta m \, g (y_2 – y_1) = -\rho \Delta V g (y_2 – y_1)$$
The change in kinetic energy of the fluid element is:
$$\Delta K = \frac{1}{2} \Delta m (v_2^2 – v_1^2) = \frac{1}{2} \rho \Delta V (v_2^2 – v_1^2)$$
Applying the Work-Energy Theorem ($W_{\text{net}} = W_{\text{pressure}} + W_{\text{gravity}} = \Delta K$):
$$(P_1 – P_2)\Delta V – \rho g (y_2 – y_1)\Delta V = \frac{1}{2}\rho (v_2^2 – v_1^2)\Delta V$$
Dividing by $\Delta V$ and grouping terms at stations $1$ and $2$:
$$P_1 + \frac{1}{2}\rho v_1^2 + \rho g y_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g y_2$$
Hence, along any given streamline:
$$\mathbf{P + \frac{1}{2}\rho v^2 + \rho g y = \text{constant}}$$
1.3 Alternative Representations & Hydraulic Heads
Depending on the analytical context in JEE problems, Bernoulli’s equation can be expressed in three distinct standard forms:
- 1. Total Energy per Unit Volume [$\text{J}/\text{m}^3 = \text{N}/\text{m}^2 = \text{Pa}$]: $$P + \frac{1}{2}\rho v^2 + \rho g y = \text{constant}$$ where $P$ is static pressure energy, $\frac{1}{2}\rho v^2$ is dynamic pressure, and $\rho g y$ is hydrostatic potential energy per unit volume.
- 2. Total Energy per Unit Mass [$\text{J}/\text{kg} = \text{m}^2/\text{s}^2$]: $$\frac{P}{\rho} + \frac{1}{2}v^2 + g y = \text{constant}$$
- 3. Total Head Form (Energy per Unit Weight) [Length, $\text{meters}$]:
$$\frac{P}{\rho g} + \frac{v^2}{2g} + y = H = \text{constant}$$
where:
- $\frac{P}{\rho g}$ = Pressure Head
- $\frac{v^2}{2g}$ = Velocity (Dynamic) Head
- $y$ = Elevation (Potential / Datum) Head
- $H$ = Total Hydraulic Head
1.4 Key Applications of Bernoulli’s Principle
A. The Venturimeter
A Venturimeter is a device used to measure the volumetric discharge rate $Q$ of liquid flowing through a pipeline. It consists of a converging inlet of cross-sectional area $A_1$, a constricted narrow throat of area $A_2$ ($A_2 < A_1$), and a gradual diverging diffuser.
For a horizontal meter ($y_1 = y_2$):
$$P_1 – P_2 = \frac{1}{2}\rho (v_2^2 – v_1^2)$$
Using the continuity equation $A_1 v_1 = A_2 v_2 \implies v_2 = \left(\frac{A_1}{A_2}\right) v_1$:
$$P_1 – P_2 = \frac{1}{2}\rho v_1^2 \left[\left(\frac{A_1}{A_2}\right)^2 – 1\right] = \frac{1}{2}\rho v_1^2 \left(\frac{A_1^2 – A_2^2}{A_2^2}\right)$$
If a differential U-tube manometer containing manometric fluid of density $\rho_m$ shows a level difference $h$:
$$P_1 – P_2 = (\rho_m – \rho) g h$$
Solving for the flow speed $v_1$ and total volumetric flow rate $Q = A_1 v_1$:
$$v_1 = \sqrt{\frac{2(P_1 – P_2)}{\rho \left[\left(\frac{A_1}{A_2}\right)^2 – 1\right]}} = A_2 \sqrt{\frac{2(\rho_m – \rho) g h}{\rho (A_1^2 – A_2^2)}}$$
$$Q = A_1 v_1 = A_1 A_2 \sqrt{\frac{2(\rho_m – \rho) g h}{\rho (A_1^2 – A_2^2)}}$$
B. Torricelli’s Law of Efflux
Consider an open tank of cross-sectional area $A$ containing liquid of density $\rho$ to a height $H$. A small hole of area $a$ ($a \ll A$) is punctured in the side wall at depth $h$ below the free surface.
Applying Bernoulli’s equation between the free liquid surface (point $1$) and the orifice exit (point $2$), both open to atmospheric pressure $P_0$:
$$P_0 + \frac{1}{2}\rho v_1^2 + \rho g h = P_0 + \frac{1}{2}\rho v_2^2 + 0$$
From continuity, $v_1 = \frac{a}{A} v_2$. When $a \ll A$, $v_1 \approx 0$, yielding Torricelli’s speed of efflux:
$$v_2 = \sqrt{2gh}$$
If the finite ratio $a/A$ is taken into account without neglecting surface velocity:
$$v_2 = \sqrt{\frac{2gh}{1 – (a/A)^2}}$$
C. Atomizer / Spray Pump / Aspirator
In a perfume spray pump, paint atomizer, or carburetor, air is driven at high speed $v_{\text{air}}$ through a narrow constriction situated directly over a vertical capillary tube dipped into liquid of density $\rho_L$.
- At the wide air chamber (stagnation), air speed is negligible and pressure is atmospheric $P_0$.
- At the narrow constriction, the high speed $v_{\text{air}}$ creates a localized low-pressure zone $P_{\text{nozzle}}$: $$P_0 – P_{\text{nozzle}} = \frac{1}{2}\rho_{\text{air}} v_{\text{air}}^2$$
- The free surface of the liquid in the reservoir is exposed to atmospheric pressure $P_0$. The pressure imbalance forces the liquid column to rise through height $h$: $$P_0 – P_{\text{nozzle}} \ge \rho_L g h \implies \frac{1}{2}\rho_{\text{air}} v_{\text{air}}^2 \ge \rho_L g h$$
- The minimum air velocity required to lift the liquid to the nozzle is: $$v_{\text{air, min}} = \sqrt{\frac{2 \rho_L g h}{\rho_{\text{air}}}}$$
- Upon reaching the high-speed air stream, the liquid stream is violently sheared and atomized into a fine aerosol mist.
D. Dynamic Aerodynamic Lift & Magnus Effect
When an aircraft wing (aerofoil) moves through air, the upper surface is curved while the lower surface is relatively flat. Streamlines are crowded above the wing, producing higher flow velocity ($v_{\text{top}} > v_{\text{bottom}}$). By Bernoulli’s equation, $P_{\text{bottom}} > P_{\text{top}}$, yielding a net upward aerodynamic lift force $F_L = (P_{\\text{bottom}} – P_{\text{top}}) A$.
Similarly, for a spinning ball (Magnus Effect), viscous boundary drag drags air along the direction of rotation. Air accelerates on one side and decelerates on the opposite side, producing a lateral pressure gradient that curves the trajectory in flight.
1.5 Critical Conceptual Insights & JEE Traps
- Trap 1: “High speed always means low pressure.” This statement is valid only along horizontal streamlines where gravitational potential energy $\rho g y$ is constant. If elevation increases, pressure can drop even if speed remains constant or decreases!
- Trap 2: Rotational vs Irrotational Flow: In irrotational flow, $\text{curl}(\vec{v}) = 0$, and the Bernoulli constant $C$ is universally identical everywhere in the fluid. In rotational flow (e.g., vortex flow), $C$ changes from one streamline to another.
- Trap 3: Unsteady Flow Pitfall: Bernoulli’s standard algebraic form cannot be applied directly to unsteady flows where $\frac{\partial \vec{v}}{\partial t} \neq 0$. The unsteady Bernoulli equation requires an acceleration potential integral: $$\int_1^2 \frac{\partial \vec{v}}{\partial t} \cdot d\vec{r} + \left(P + \frac{1}{2}\rho v^2 + \rho g y\right)_1 = \left(P + \frac{1}{2}\rho v^2 + \rho g y\right)_2$$
- Trap 4: Cavitation & Absolute vs Gauge Pressure: Absolute pressure in an ordinary liquid cannot drop below its saturated vapor pressure ($P \ge P_{\text{vap}}$). If Bernoulli calculations yield $P < P_{\text{vap}}$, liquid boils spontaneously at ambient temperature, producing vapor bubbles (cavitation) that collapse violently and disrupt flow.
2. Four Graded Solved Examples
Example 1: Flow Rate Measurement in a Horizontal Venturimeter (JEE Main Level)
Problem Statement: Water (density $\rho_w = 1000\text{ kg/m}^3$) flows through a horizontal pipeline of internal diameter $D_1 = 20\text{ cm}$. A Venturimeter with a constricted throat diameter $D_2 = 10\text{ cm}$ is connected into the line. A differential mercury manometer ($\rho_m = 13600\text{ kg/m}^3$) connected between the inlet and throat indicates a mercury column height difference $h = 10\text{ cm}$. Taking $g = 9.8\text{ m/s}^2$, determine:
- (a) The pressure difference $\Delta P = P_1 – P_2$ between the inlet and the throat.
- (b) The flow velocity $v_1$ in the main pipe.
- (c) The volumetric discharge rate $Q$ in liters per second ($\text{L/s}$).
Step-by-Step Solution:
Step 1: Calculate cross-sectional areas and area ratio:
$$A_1 = \frac{\pi D_1^2}{4} = \frac{\pi (0.20)^2}{4} = 0.01\pi \text{ m}^2$$
$$A_2 = \frac{\pi D_2^2}{4} = \frac{\pi (0.10)^2}{4} = 0.0025\pi \text{ m}^2$$
The area ratio is $\frac{A_1}{A_2} = \left(\frac{D_1}{D_2}\right)^2 = \left(\frac{20}{10}\right)^2 = 4 \implies \left(\frac{A_1}{A_2}\right)^2 = 16$.
Step 2: Calculate the pressure difference $\Delta P$:
From manometer hydrostatics:
$$\Delta P = P_1 – P_2 = (\rho_m – \rho_w) g h = (13600 – 1000) \times 9.8 \times 0.10$$
$$\Delta P = 12600 \times 0.98 = 12348 \text{ N/m}^2 = 12.35 \text{ kPa}$$
Step 3: Calculate flow velocity $v_1$ using Bernoulli’s equation:
$$P_1 – P_2 = \frac{1}{2}\rho_w v_1^2 \left[\left(\frac{A_1}{A_2}\right)^2 – 1\right]$$
$$12348 = \frac{1}{2}(1000) v_1^2 [16 – 1] = 500 \times 15 \times v_1^2 = 7500 v_1^2$$
$$v_1^2 = \frac{12348}{7500} = 1.6464 \implies v_1 = \sqrt{1.6464} \approx 1.283 \text{ m/s}$$
Step 4: Calculate the volumetric flow rate $Q$:
$$Q = A_1 v_1 = (0.01\pi \text{ m}^2) \times 1.2831 \text{ m/s} = 0.04031 \text{ m}^3\text{/s}$$
Converting to liters per second ($1\text{ m}^3 = 1000\text{ L}$):
$$Q = 0.04031 \times 1000 \approx 40.31 \text{ L/s}$$
Key Takeaway: In differential manometry, the effective driving head is $(\rho_m – \rho)gh$, NOT simply $\rho_m gh$, because fluid in the manometer legs balances part of the liquid column.
Example 2: Time to Empty a Cylindrical Vessel via Bottom Orifice (Calculus & Torricelli)
Problem Statement: A large vertical cylindrical water tank of radius $R = 1.0\text{ m}$ is filled with water to an initial height $H = 4.9\text{ m}$. A circular drain orifice of radius $r = 1.0\text{ cm}$ is opened at the center of the flat bottom. Assuming quasi-steady frictionless flow and taking $g = 9.8\text{ m/s}^2$, calculate:
- (a) The time $T$ required to completely drain the tank.
- (b) The time $t_{1/2}$ required for the water level to drop from $H$ to $\frac{H}{2}$.
Step-by-Step Solution:
Step 1: Set up the differential equation from continuity and Torricelli’s law:
Let $y(t)$ be the instantaneous height of the free water surface above the orifice. The cross-sectional area of the tank is $A = \pi R^2$, and the orifice area is $a = \pi r^2$.
The instantaneous rate of change of fluid volume is:
$$\frac{dV}{dt} = A \frac{dy}{dt} = -a v_{\text{efflux}} = -a \sqrt{2gy}$$
$$\frac{dy}{dt} = -\left(\frac{a}{A}\right)\sqrt{2g} \, y^{1/2} = -\left(\frac{r}{R}\right)^2 \sqrt{2g} \, y^{1/2}$$
Step 2: Separate variables and integrate:
$$\int_H^0 y^{-1/2}\, dy = -\left(\frac{r}{R}\right)^2 \sqrt{2g} \int_0^T dt$$
$$\left[ 2y^{1/2} \right]_H^0 = -2\sqrt{H} = -\left(\frac{r}{R}\right)^2 \sqrt{2g}\, T$$
$$T = \left(\frac{R}{r}\right)^2 \sqrt{\frac{2H}{g}}$$
Step 3: Substitute numerical values for total draining time $T$:
$$\frac{R}{r} = \frac{1.0\text{ m}}{0.01\text{ m}} = 100 \implies \left(\frac{R}{r}\right)^2 = 10000$$
$$\sqrt{\frac{2H}{g}} = \sqrt{\frac{2 \times 4.9}{9.8}} = \sqrt{1.0} = 1.0\text{ s}$$
$$T = 10000 \times 1.0\text{ s} = 10000\text{ s} \approx 2\text{ hr } 46\text{ min } 40\text{ s}$$
Step 4: Calculate time $t_{1/2}$ for height to fall to $H/2$:
$$\int_H^{H/2} y^{-1/2}\, dy = 2\left(\sqrt{H/2} – \sqrt{H}\right) = -2\sqrt{H}\left(1 – \frac{1}{\sqrt{2}}\right)$$
$$t_{1/2} = T \left(1 – \frac{1}{\sqrt{2}}\right) = 10000 \times (1 – 0.7071) = 2929\text{ s}$$
Key Takeaway: Emptying the first half of the tank takes only $29.3\%$ of the total time ($2929\text{ s}$), while emptying the remaining half takes $70.7\%$ of the time ($7071\text{ s}$) because the hydrostatic driving head continually decreases.
Example 3: Siphon Dynamics & Cavitation Threshold (JEE Advanced Level)
Problem Statement: A siphon tube of uniform internal cross-sectional area is employed to transfer water ($\rho = 1000\text{ kg/m}^3$) from an elevated reservoir open to atmosphere ($P_0 = 1.01 \times 10^5\text{ N/m}^2$). The crest of the siphon is at height $h_1 = 3.0\text{ m}$ above the reservoir water surface, and the discharge outlet opens to atmosphere at depth $h_2 = 4.0\text{ m}$ below the reservoir surface. Taking $g = 10\text{ m/s}^2$ and assuming frictionless flow:
- (a) Determine the efflux speed $v$ at the siphon discharge exit.
- (b) Calculate the absolute pressure $P_{\text{crest}}$ at the highest point (summit) of the siphon.
- (c) If the vapor pressure of water at the operating temperature is $P_{\text{vap}} = 2.0 \times 10^3\text{ Pa}$, find the theoretical maximum allowable height $h_{1,\text{max}}$ of the crest above the reservoir surface before cavitation occurs.
Step-by-Step Solution:
Step 1: Efflux velocity at discharge outlet:
Choose the reservoir surface as datum $y = 0$, where $P = P_0$ and $v_{\text{surface}} \approx 0$. Applying Bernoulli’s equation between the reservoir surface and the discharge exit ($y = -h_2$, $P = P_0$):
$$P_0 + 0 + 0 = P_0 + \frac{1}{2}\rho v^2 + \rho g (-h_2)$$
$$v = \sqrt{2g h_2} = \sqrt{2 \times 10 \times 4.0} = \sqrt{80} \approx 8.944 \text{ m/s}$$
Step 2: Pressure at the crest of the siphon:
Because the siphon tube has uniform cross-section, by continuity the flow speed at the crest is identical to the discharge speed: $v_{\text{crest}} = v$.
Applying Bernoulli’s equation between reservoir surface ($y = 0$) and the crest ($y = +h_1$):
$$P_0 + 0 + 0 = P_{\text{crest}} + \frac{1}{2}\rho v^2 + \rho g h_1$$
Since $\frac{1}{2}\rho v^2 = \rho g h_2$ from Step 1:
$$P_{\text{crest}} = P_0 – \rho g h_1 – \frac{1}{2}\rho v^2 = P_0 – \rho g (h_1 + h_2)$$
Substitute numerical values:
$$P_{\text{crest}} = 1.01 \times 10^5 – 1000 \times 10 \times (3.0 + 4.0) = 101000 – 70000 = 31000 \text{ Pa} = 31.0 \text{ kPa}$$
Step 3: Maximum allowable crest height $h_{1,\text{max}}$ to prevent cavitation:
Flow continuity breaks down if $P_{\text{crest}} < P_{\text{vap}}$. The critical threshold is:
$$P_{\text{crest}} \ge P_{\text{vap}} \implies P_0 – \rho g (h_{1,\text{max}} + h_2) = P_{\text{vap}}$$
$$h_{1,\text{max}} + h_2 = \frac{P_0 – P_{\text{vap}}}{\rho g} = \frac{101000 – 2000}{1000 \times 10} = \frac{99000}{10000} = 9.9 \text{ m}$$
$$h_{1,\text{max}} = 9.9 – h_2 = 9.9 – 4.0 = 5.9 \text{ m}$$
Key Takeaway: The driving force of a siphon is the total height difference $h_2$ between the reservoir surface and the discharge exit. However, the absolute summit pressure drops below atmospheric by $\rho g(h_1 + h_2)$, establishing a rigorous barometric ceiling on siphon height.
Example 4: Quantitative Dynamics of a Perfume Atomizer / Spray Pump
Problem Statement: A laboratory spray atomizer consists of a horizontal cylinder of diameter $D = 2.0\text{ cm}$ fitted with a piston that forces air through a constricted horizontal nozzle of diameter $d = 2.0\text{ mm}$. A vertical dip tube of length $h = 10\text{ cm}$ has its upper tip opening flush into the nozzle constriction, while its bottom end dips into an open container of liquid (density $\rho_L = 800\text{ kg/m}^3$). Taking air density $\rho_{\text{air}} = 1.0\text{ kg/m}^3$ and $g = 10\text{ m/s}^2$:
- (a) Calculate the minimum air speed $v_{\text{nozzle}}$ required at the nozzle constriction to draw liquid up to the nozzle level.
- (b) Calculate the minimum forward speed $u_{\text{piston}}$ with which the user must depress the piston to initiate liquid suction.
Step-by-Step Solution:
Step 1: Calculate the pressure drop required for liquid lift:
The liquid reservoir is exposed to atmospheric pressure $P_0$. For liquid to rise to height $h = 0.10\text{ m}$, the pressure $P_{\text{nozzle}}$ at the nozzle must satisfy:
$$P_0 – P_{\text{nozzle}} \ge \rho_L g h = 800 \times 10 \times 0.10 = 800 \text{ Pa}$$
Step 2: Relate pressure drop to air velocity via Bernoulli’s equation:
Applying Bernoulli’s equation to the air stream between the main cylinder (where speed is $u_{\text{piston}}$) and the nozzle exit (where speed is $v_{\text{nozzle}}$ and pressure is $P_{\text{nozzle}}$):
$$P_{\text{cyl}} + \frac{1}{2}\rho_{\text{air}} u_{\text{piston}}^2 = P_{\text{nozzle}} + \frac{1}{2}\rho_{\text{air}} v_{\text{nozzle}}^2$$
Because $D/d = 20\text{ mm} / 2\text{ mm} = 10$, by continuity $u_{\text{piston}} = (d/D)^2 v_{\text{nozzle}} = \frac{1}{100} v_{\text{nozzle}}$. Therefore, $u_{\text{piston}}^2 = 10^{-4} v_{\text{nozzle}}^2 \approx 0$. Since the air cylinder is driven from atmospheric stagnation ($P_{\text{cyl}} \approx P_0$):
$$P_0 – P_{\text{nozzle}} = \frac{1}{2}\rho_{\text{air}} v_{\text{nozzle}}^2$$
Step 3: Solve for the minimum air speed $v_{\text{nozzle}}$:
$$\frac{1}{2} (1.0) v_{\text{nozzle}}^2 = 800 \implies v_{\text{nozzle}}^2 = 1600$$
$$v_{\text{nozzle}} = 40.0 \text{ m/s}$$
Step 4: Solve for minimum piston speed $u_{\text{piston}}$:
$$u_{\text{piston}} = \left(\frac{d}{D}\right)^2 v_{\text{nozzle}} = \left(\frac{1}{10}\right)^2 \times 40.0 = \frac{40.0}{100} = 0.40 \text{ m/s} = 40 \text{ cm/s}$$
Key Takeaway: High-speed air creates a strong localized suction depression via the dynamic pressure term $\frac{1}{2}\rho_{\text{air}} v^2$. The large area constriction ratio $(D/d)^2 = 100$ allows modest hand piston motion ($40\text{ cm/s}$) to generate a $40\text{ m/s}$ high-speed jet capable of atomizing dense liquid.
3. Practice Worksheet (10 Questions)
Questions 1 to 5: JEE Main Level (Single Correct Option)
Q1. In the head form of Bernoulli’s equation, $\frac{P}{\rho g} + \frac{v^2}{2g} + y = \text{constant}$, the physical dimensions of each individual term are identical to that of:
- (A) Energy $[\text{M}^1 \text{L}^2 \text{T}^{-2}]$
- (B) Pressure $[\text{M}^1 \text{L}^{-1} \text{T}^{-2}]$
- (C) Length $[\text{M}^0 \text{L}^1 \text{T}^0]$
- (D) Power $[\text{M}^1 \text{L}^2 \text{T}^{-3}]$
Q2. Water ($\rho = 1000\text{ kg/m}^3$) flows steadily through a horizontal pipe. At section $1$, the flow velocity is $v_1 = 2\text{ m/s}$ and the static pressure is $P_1 = 300\text{ kPa}$. The pipe then tapers to section $2$ where the diameter is halved. Neglecting viscous friction, the static pressure $P_2$ at section $2$ is:
- (A) $270\text{ kPa}$
- (B) $240\text{ kPa}$
- (C) $285\text{ kPa}$
- (D) $180\text{ kPa}$
Q3. A horizontal Venturimeter has inlet area $A_1$ and throat area $A_2$ such that $A_1 = 2 A_2$. Two vertical open piezometer tubes connected to the inlet and throat show a liquid column height difference of $h$. The velocity of approach $v_1$ in the main pipe is given by:
- (A) $\sqrt{\frac{2gh}{3}}$
- (B) $\sqrt{\frac{gh}{3}}$
- (C) $\sqrt{2gh}$
- (D) $2\sqrt{gh}$
Q4. A closed cylindrical storage tank containing liquid of density $\rho$ has compressed air at pressure $P$ maintained above the liquid surface. A small orifice is punctured in the sidewall at depth $h$ below the liquid surface. The speed of efflux of liquid discharging into ambient atmosphere (pressure $P_0$) is:
- (A) $\sqrt{2gh}$
- (B) $\sqrt{\frac{2(P – P_0)}{\rho} + 2gh}$
- (C) $\sqrt{\frac{2(P + P_0)}{\rho} + 2gh}$
- (D) $\sqrt{\frac{P – P_0}{\rho} + gh}$
Q5. An atomizer uses a horizontal stream of air (density $\rho_a = 1.0\text{ kg/m}^3$) blown across the top of a vertical tube dipped into liquid (density $\rho_L = 800\text{ kg/m}^3$). If the top of the tube is at height $h = 10\text{ cm}$ above the liquid surface and $g = 10\text{ m/s}^2$, the minimum air speed required to draw liquid to the top of the tube is:
- (A) $20\text{ m/s}$
- (B) $28.3\text{ m/s}$
- (C) $40\text{ m/s}$
- (D) $56.6\text{ m/s}$
Questions 6 to 8: JEE Advanced Level (One or More than One Correct)
Q6. Which of the following statements regarding Bernoulli’s principle and fluid flow fields are correct?
- (A) Bernoulli’s equation is fundamentally a restatement of the Work-Energy Theorem for continuous ideal fluid systems.
- (B) In steady irrotational flow ($\nabla \times \vec{v} = 0$), the Bernoulli constant has the exact same numerical value for all streamlines throughout the fluid.
- (C) For rotational flow, the Bernoulli equation is valid across different streamlines provided viscous dissipation is absent.
- (D) In a horizontal streamline, regions of higher velocity correspond strictly to regions of lower static pressure.
Q7. A siphon tube of uniform cross-section transfers liquid from an open container. The highest point (summit) of the siphon is at height $h_1$ above the liquid surface, and the outlet is at depth $h_2$ below the surface. Which of the following statements are correct?
- (A) The flow rate depends directly on the vertical distance $h_2$ between the liquid surface and the discharge outlet.
- (B) The static pressure in the liquid reaches its global minimum at the summit of the siphon.
- (C) The theoretical maximum height $h_1$ is constrained by the condition that pressure at the summit cannot drop below the liquid’s saturated vapor pressure.
- (D) If the discharge outlet is raised above the container surface ($h_2 < 0$), liquid continues to flow out due to siphon momentum.
Q8. A tall cylindrical container of height $H$ completely filled with water stands on a horizontal floor. Two small identical holes are punctured in its vertical wall at depths $h_1$ and $h_2$ from the top surface. Let $R_1$ and $R_2$ be the horizontal ranges of the emerging water jets on the floor. Which of the following are correct?
- (A) The horizontal range of a jet emerging from depth $h$ is $R = 2\sqrt{h(H – h)}$.
- (B) The horizontal ranges $R_1$ and $R_2$ are equal if $h_1 + h_2 = H$.
- (C) The maximum possible horizontal range is $R_{\text{max}} = H$, achieved when the hole is located at depth $h = H/2$.
- (D) The speed of efflux is maximum for a hole punched at depth $h = H/2$.
Questions 9 to 10: Numerical / Integer Answer Type
Q9. A large cylindrical open water tank has a cross-sectional area $A = 2.0\text{ m}^2$. A small circular hole of area $a = 2.0\text{ cm}^2$ is opened at the bottom of the tank. Initially, the water level is $H_1 = 5.0\text{ m}$. Taking $g = 10\text{ m/s}^2$ and assuming quasi-steady frictionless flow, find the time (in seconds) required for the water level to drop from $5.0\text{ m}$ to $1.25\text{ m}$.
Q10. Water (density $\rho = 1000\text{ kg/m}^3$) flows through a horizontal constriction where the cross-sectional area reduces from $A_1 = 0.05\text{ m}^2$ to $A_2 = 0.025\text{ m}^2$. The static pressure difference measured between sections $1$ and $2$ is $\Delta P = P_1 – P_2 = 6000\text{ Pa}$. Calculate the volumetric flow rate of water $Q$ through the pipe in liters per second ($\text{L/s}$).
4. Detailed Solutions & Answer Key
Quick Answer Key
| Question | Correct Answer | Topic / Category |
|---|---|---|
| Q1 | (C) | Dimensions of Hydraulic Heads |
| Q2 | (A) | Bernoulli & Continuity in Tapered Pipe |
| Q3 | (A) | Venturimeter Piezometer Relation |
| Q4 | (B) | Torricelli Efflux with Pressurized Gas |
| Q5 | (C) | Atomizer Suction Velocity Threshold |
| Q6 | (A), (B), (D) | Theoretical Foundations of Bernoulli |
| Q7 | (A), (B), (C) | Siphon Mechanics & Cavitation Limit |
| Q8 | (A), (B), (C) | Torricelli Jet Range & Symmetry |
| Q9 | 5000 | Tank Drainage Time Calculation |
| Q10 | 100 | Volumetric Flow Rate Discharge |
Step-by-Step Solutions
Solution 1:
In the head form: $\frac{P}{\rho g} + \frac{v^2}{2g} + y = \text{constant}$.
- Elevation head $y$ has dimensions of length: $[\text{M}^0 \text{L}^1 \text{T}^0]$.
- Velocity head: $\left[\frac{v^2}{2g}\right] = \frac{(\text{L}\text{T}^{-1})^2}{\text{L}\text{T}^{-2}} = \frac{\text{L}^2 \text{T}^{-2}}{\text{L}\text{T}^{-2}} = [\text{L}]$.
- Pressure head: $\left[\frac{P}{\rho g}\right] = \frac{\text{M}\text{L}^{-1}\text{T}^{-2}}{(\text{M}\text{L}^{-3})(\text{L}\text{T}^{-2})} = \frac{\text{M}\text{L}^{-1}\text{T}^{-2}}{\text{M}\text{L}^{-2}\text{T}^{-2}} = [\text{L}]$.
Every term has the dimension of length. Correct Option: (C).
Solution 2:
Since the diameter is halved ($D_2 = D_1 / 2$), the area reduces by a factor of $4$: $A_2 = A_1 / 4$.
By continuity: $A_1 v_1 = A_2 v_2 \implies v_2 = 4 v_1 = 4 \times 2 = 8\text{ m/s}$.
Applying Bernoulli’s equation for a horizontal pipe:
$$P_2 = P_1 – \frac{1}{2}\rho (v_2^2 – v_1^2) = 300\text{ kPa} – \frac{1}{2}(1000)(8^2 – 2^2)\text{ Pa}$$
$$P_2 = 300000 – 500 \times (64 – 4) = 300000 – 30000 = 270000\text{ Pa} = 270\text{ kPa}$$
Correct Option: (A).
Solution 3:
By continuity, $v_2 = \frac{A_1}{A_2} v_1 = 2 v_1$.
The piezometer height difference gives the static pressure difference:
$$P_1 – P_2 = \rho g h$$
From Bernoulli’s equation:
$$P_1 – P_2 = \frac{1}{2}\rho (v_2^2 – v_1^2) = \frac{1}{2}\rho [(2v_1)^2 – v_1^2] = \frac{3}{2}\rho v_1^2$$
$$\rho g h = \frac{3}{2}\rho v_1^2 \implies v_1^2 = \frac{2gh}{3} \implies v_1 = \sqrt{\frac{2gh}{3}}$$
Correct Option: (A).
Solution 4:
Apply Bernoulli’s equation between the pressurized surface (point $1$) and the exit orifice (point $2$):
$$P + 0 + \rho g h = P_0 + \frac{1}{2}\rho v^2 + 0$$
$$\frac{1}{2}\rho v^2 = (P – P_0) + \rho g h \implies v = \sqrt{\frac{2(P – P_0)}{\rho} + 2gh}$$
Correct Option: (B).
Solution 5:
To lift a liquid column of height $h = 0.10\text{ m}$, the suction pressure required is:
$$\Delta P = \rho_L g h = 800 \times 10 \times 0.10 = 800\text{ Pa}$$
From the high-speed air stream above the tube:
$$\Delta P = \frac{1}{2}\rho_a v^2 \implies 800 = \frac{1}{2}(1.0) v^2 \implies v^2 = 1600 \implies v = 40\text{ m/s}$$
Correct Option: (C).
Solution 6:
- (A) is Correct: The equation is obtained directly by integrating $\vec{F} = m\vec{a}$ along a streamline, which is the Work-Energy Theorem for fluid continuum.
- (B) is Correct: For irrotational flow ($\nabla \times \vec{v} = 0$), Euler’s equation integrates globally, so the constant is identical throughout the entire flow field.
- (C) is Incorrect: In rotational flow, the Bernoulli constant differs from one streamline to another; it cannot be applied across different streamlines.
- (D) is Correct: Along a horizontal streamline ($y = \text{constant}$), $P + \frac{1}{2}\rho v^2 = \text{constant}$, meaning higher velocity strictly corresponds to lower pressure.
Correct Options: (A), (B), (D).
Solution 7:
- (A) is Correct: The efflux velocity is $v = \sqrt{2g h_2}$, directly governed by the elevation drop between reservoir surface and exit.
- (B) is Correct: At height $y$, $P = P_0 – \rho g y – \frac{1}{2}\rho v^2$. Since $v$ is uniform throughout the tube, $P$ is lowest where $y$ is highest (at the summit $h_1$).
- (C) is Correct: If $P_{\text{summit}} < P_{\text{vap}}$, liquid cavitates and vapor locks the tube, halting siphon action.
- (D) is Incorrect: If $h_2 < 0$ (outlet above reservoir), pressure at the exit would have to exceed atmospheric for outward flow, which is impossible; fluid flows back into the container.
Correct Options: (A), (B), (C).
Solution 8:
- Efflux speed at depth $h$: $v = \sqrt{2gh}$.
- Vertical height of fall to floor: $y = H – h$.
- Time of flight: $t = \sqrt{\frac{2(H – h)}{g}}$.
- Horizontal range: $R = v \cdot t = \sqrt{2gh} \sqrt{\frac{2(H – h)}{g}} = 2\sqrt{h(H – h)}$. Hence (A) is Correct.
- For $h_1$ and $h_2 = H – h_1$: $R_2 = 2\sqrt{(H – h_1)(H – (H – h_1))} = 2\sqrt{(H – h_1)h_1} = R_1$. Hence (B) is Correct.
- Differentiating $R(h)$ with respect to $h$ gives maximum at $h = H/2$, with $R_{\text{max}} = 2\sqrt{\frac{H}{2} \cdot \frac{H}{2}} = H$. Hence (C) is Correct.
- Efflux speed $v = \sqrt{2gh}$ increases monotonically with $h$; it is maximum at the bottom ($h = H$), not at $h = H/2$. Hence (D) is Incorrect.
Correct Options: (A), (B), (C).
Solution 9:
From mass conservation: $A \left(-\frac{dy}{dt}\right) = a \sqrt{2gy}$.
$$\Delta t = \frac{A}{a} \sqrt{\frac{2}{g}} \int_{H_2}^{H_1} \frac{dy}{2\sqrt{y}} = \frac{A}{a} \sqrt{\frac{2}{g}} (\sqrt{H_1} – \sqrt{H_2})$$
Given:
- $A = 2.0\text{ m}^2$, $a = 2.0\text{ cm}^2 = 2.0 \times 10^{-4}\text{ m}^2 \implies \frac{A}{a} = \frac{2.0}{2.0 \times 10^{-4}} = 10^4$.
- $g = 10\text{ m/s}^2 \implies \sqrt{\frac{2}{g}} = \sqrt{\frac{2}{10}} = \frac{1}{\sqrt{5}}$.
- $H_1 = 5.0\text{ m} \implies \sqrt{H_1} = \sqrt{5}$.
- $H_2 = 1.25\text{ m} = \frac{5}{4}\text{ m} \implies \sqrt{H_2} = \frac{\sqrt{5}}{2}$.
$$\sqrt{H_1} – \sqrt{H_2} = \sqrt{5} – \frac{\sqrt{5}}{2} = \frac{\sqrt{5}}{2}$$
$$\Delta t = 10^4 \times \frac{1}{\sqrt{5}} \times \frac{\sqrt{5}}{2} = \frac{10^4}{2} = 5000\text{ s}$$
Final Answer: 5000
Solution 10:
Given $A_1 = 0.05\text{ m}^2$ and $A_2 = 0.025\text{ m}^2 \implies \frac{A_1}{A_2} = 2$.
From continuity, $v_2 = 2 v_1$.
Applying Bernoulli’s equation for a horizontal streamline:
$$\Delta P = P_1 – P_2 = \frac{1}{2}\rho (v_2^2 – v_1^2) = \frac{1}{2}\rho (4 v_1^2 – v_1^2) = \frac{3}{2}\rho v_1^2$$
$$6000 = \frac{3}{2}(1000) v_1^2 = 1500 v_1^2 \implies v_1^2 = \frac{6000}{1500} = 4 \implies v_1 = 2.0\text{ m/s}$$
The volumetric discharge rate is:
$$Q = A_1 v_1 = 0.05\text{ m}^2 \times 2.0\text{ m/s} = 0.10\text{ m}^3\text{/s}$$
Converting to liters per second ($1\text{ m}^3 = 1000\text{ L}$):
$$Q = 0.10 \times 1000 = 100\text{ L/s}$$
Final Answer: 100