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A pipe of cross-sectional area $A_1 = 8\text{ cm}^2$ carries fluid at velocity $v_1 = 50\text{ cm/s}$. The pipe narrows to $A_2 = 2\text{ cm}^2$. The flow rate through the pipe is:

R
Solution written and verified by Roshan, science educator with 5 years of experience teaching NEET and JEE aspirants. Last reviewed September 2026.
Options
1
$200$ cm$^3$/s
2
$400$ cm$^3$/s
3
$100$ cm$^3$/s
4
$800$ cm$^3$/s
Correct Answer
$400$ cm$^3$/s
Solution
1

Flow rate $Q = A_1 v_1 = 8\text{ cm}^2 \times 50\text{ cm/s}$

2

$Q = 400\text{ cm}^3/\text{s}$

Answer: 400 cm$^3$/s

$Q = Av$ = volumetric flow rate (constant for incompressible fluid)
$Q = 8\times50 = 400$ cm$^3$/s
Theory: Fluid Mechanics
1. Fluid Mechanics Fundamentals

Ideal fluid: incompressible, non-viscous, irrotational (no turbulence). Real fluids: have viscosity, compressibility at high speeds. Streamline: curve tangent to velocity vector at each point; no two streamlines cross. Stream tube: bundle of streamlines. Stagnation point: where fluid velocity = 0; full $\frac{1}{2}\rho v^2$ converted to pressure ($P_0 = P + \frac{1}{2}\rho v^2$). Pitot tube: measures stagnation pressure to find fluid speed.

2. Bernoulli's Principle and Applications

$P + \frac{1}{2}\rho v^2 + \rho gh = $ const along streamline. Applications: Aeroplane wings: faster air flow over curved upper surface → lower pressure above wing than below → net upward force (lift). Spinning ball (Magnus effect): cricket/football spin causes curved trajectory due to pressure difference from asymmetric airflow. Chimney effect: natural ventilation by warm rising air. Bunsen burner: gas flow entrains air for combustion.

3. Viscosity and Poiseuille's Law

Viscous flow in pipe (Poiseuille\'s law): $Q = \frac{\pi r^4 \Delta P}{8\eta L}$ (flow rate proportional to $r^4$, inverse to viscosity $\eta$ and length $L$). Doubling radius → 16× flow rate. This is why arterial narrowing (atherosclerosis) dramatically reduces blood flow. Blood flow: blood is non-Newtonian fluid (viscosity depends on shear rate). Stokes drag: $F = 6\pi\eta rv$ for sphere of radius $r$ at speed $v$. Terminal velocity: $v_t = 2r^2g(\rho_{sphere}-\rho_{fluid})/9\eta$.

4. Surface Tension Phenomena

Surface tension $T$: force per unit length = $F/l$ (N/m). Energy per unit area: $T$ (J/m$^2$). Pressure excess inside bubble: soap bubble = $4T/r$ (two surfaces); liquid drop = $2T/r$; air bubble in liquid = $2T/r$. Capillary rise: $h = 2T\cos\theta/\rho g r$. Meniscus: concave (wetting liquids, $\theta<90°$, like water in glass). Convex: non-wetting ($\theta>90°$, mercury in glass). Detergents reduce surface tension → easier cleaning.

5. Pressure in a Fluid at Rest

Pressure at depth h below a free surface is $P = P_0 + \rho gh$, and it depends only on the depth, not on the shape or total volume of the container — the hydrostatic paradox. Pascal's law follows: pressure applied to an enclosed fluid is transmitted undiminished throughout, which is the working principle of the hydraulic lift, where a small force on a small piston produces a large force on a large one in the ratio of their areas.

6. Archimedes' Principle and Floating Bodies

A body immersed in a fluid experiences an upward buoyant force equal to the weight of fluid displaced. It floats when its average density is less than the fluid's, and the fraction submerged equals the ratio of the two densities — which is why roughly nine-tenths of an iceberg lies below water. Apparent weight is the true weight minus the buoyant force, and a body of exactly the fluid's density hangs suspended at any depth.

7. Equation of Continuity

For an incompressible fluid in steady flow, $A_1v_1 = A_2v_2$, so the flow speeds up where the pipe narrows. This is conservation of mass rather than energy, and it is usually the first equation to write in a flow problem — it supplies the relation between the two speeds that Bernoulli's equation then needs. It also explains why a stream of water from a tap narrows as it falls: gravity increases v, so A must decrease.

Where students lose the mark

Thinking pressure depends on the container's width. A narrow tube and a wide tank of the same depth have identical pressure at the bottom.

Forgetting atmospheric pressure. Gauge pressure is $\rho gh$; absolute pressure adds $P_0$. Check which the question wants before answering.

Frequently Asked Questions
1. What is the equation of continuity? ⌄
For incompressible fluid: $A_1v_1 = A_2v_2 = Q$ (constant volumetric flow rate). For compressible fluid: $\rho_1 A_1 v_1 = \rho_2 A_2 v_2$ (mass flow rate constant). Narrower section → higher speed (fluid must flow faster through smaller area to maintain same flow rate).
2. How is Bernoulli's equation applied? ⌄
$P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2$. For horizontal pipe ($h_1=h_2$): $P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$. Since $v_2 > v_1$ (narrower section): $P_2 < P_1$ (pressure decreases in narrow section). This is the Venturi effect.
3. What is a Venturi meter? ⌄
Device using Bernoulli + continuity to measure flow rate. Pressure difference between wide and narrow sections: $\Delta P = P_1-P_2 = \frac{1}{2}\rho(v_2^2-v_1^2)$. Using continuity $v_2 = A_1v_1/A_2$: $Q = A_1v_1 = A_2\sqrt{\frac{2\Delta P/\rho}{1-(A_2/A_1)^2}}$.
4. What is the Torricelli theorem? ⌄
Liquid draining from hole in tank at depth $h$ below surface: exit speed $v = \sqrt{2gh}$ (applying Bernoulli between surface and hole). Time to empty tank of height $H$ with hole area $a$ and tank area $A$: $t = \frac{A}{a}\sqrt{2H/g}$.
5. What is laminar vs turbulent flow? ⌄
Laminar: smooth, parallel streamlines. Occurs at low speeds. Reynolds number $Re = \rho vD/\eta < 2000$ (laminar). Turbulent: irregular, chaotic. $Re > 4000$ (turbulent). $2000 < Re < 4000$: transition region. Turbulence increases drag. In pipes: laminar flow has parabolic velocity profile; turbulent: flatter profile with thin laminar sublayer at wall.
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