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अध्याय 9: तरलों के यांत्रिकी गुण

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CBSE • कक्षा XI • Physics • अध्याय 9

अनुमानित समय: 45 Mins

प्रगति: अध्ययनरत

तरलों के यांत्रिकी गुण

In Class 11 Physics, "Mechanical Properties of Fluids" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

अध्याय CBT ऑनलाइन टेस्ट दें सभी अध्याय देखें

प्रस्तावना अध्याय अवधारणाएँ अध्याय सारांश स्व-मूल्यांकन अभ्यास टेस्ट दें

🌊 Have You Ever Wondered?

How can a 100-ton hydraulic excavator lift massive boulders with the push of a joystick, or how does a curved aircraft wing generate lift to hoist a 500-passenger Boeing 747 into the sky? Pascal's Law, Bernoulli's Principle, and Surface Tension govern fluid mechanics.

यह अध्याय क्यों महत्वपूर्ण है

In Class 11 Physics, "Mechanical Properties of Fluids" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

अध्ययन से पूर्व (आवश्यक ज्ञान)

Pressure and buoyant force from Class 9.
Archimedes' principle.
Conservation of energy.

इस अध्याय के लक्ष्य

State Pascal's Law and apply it to Hydraulic Lift and Hydraulic Brakes ($F_2 = F_1 \frac{A_2}{A_1}$).
Analyze variation of fluid pressure with depth ($P = P_a + \rho g h$).
State the Equation of Continuity for steady incompressible flow: $A_1 v_1 = A_2 v_2$.
State and apply Bernoulli's Principle: $P + \frac{1}{2}\rho v^2 + \rho g h = \text{constant}$ (Venturi-meter, aerodynamic lift).
Define Viscosity, Stokes' Law ($F = 6\pi\eta r v$), and Terminal Velocity ($v_t = \frac{2r^2(\rho - \sigma)g}{9\eta}$).
Analyze Surface Tension ($S = F/L$), Surface Energy, and Capillary Rise ($h = \frac{2S\cos\theta}{r\rho g}$).

अध्याय रूपरेखा एवं प्रगति

1 1. Pascal's Law & Hydraulic Machine...

2 2. Continuity & Bernoulli's Princip...

3 3. Viscosity, Stokes' Law & Surface...

सम्पूर्ण सैद्धांतिक एवं वैचारिक अध्ययन

1. Pascal's Law & Hydraulic Machines

Pascal's Law: Pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel.
• Hydraulic Lift: $$\mathbf{\frac{F_1}{A_1} = \frac{F_2}{A_2} \implies F_2 = F_1 \left(\frac{A_2}{A_1}\right)}$$ A tiny force on small piston $A_1$ lifts a massive vehicle on large piston $A_2$!

2. Continuity & Bernoulli's Principle

Equation of Continuity: Mass conservation in fluid stream: $\mathbf{A_1 v_1 = A_2 v_2 = \text{constant}}$ (Water rushes faster through a constricted nozzle).
Bernoulli's Theorem: For streamline flow of an ideal fluid, total energy per unit volume is constant: $$\mathbf{P + \frac{1}{2}\rho v^2 + \rho g h = \text{constant}}$$ Aerodynamic Lift: Air travels faster over curved top of wing → lower pressure on top → upward lift force!

3. Viscosity, Stokes' Law & Surface Tension

Stokes' Law & Terminal Velocity: Drag on falling sphere: $F = 6\pi\eta r v$. Balances gravity to yield constant Terminal Velocity: $$\mathbf{v_t = \frac{2r^2(\rho - \sigma)g}{9\eta}} \quad (\text{Why raindrops don't accelerate to supersonic speeds!})$$
Capillary Rise: Liquid rises in narrow glass tube: $\mathbf{h = \frac{2S\cos\theta}{r\rho g}}$ (nourishing sap in tree roots).

Mechanical Properties of Fluids - Key Conceptual & Analytical Model

अध्याय का सार संक्षेप एवं 10 मुख्य निष्कर्ष

मुख्य बिंदु 1

Pascal's Principle: Undiminished pressure transmission enabling hydraulic multiplication.

मुख्य बिंदु 2

Equation of Continuity: Conservation of fluid mass flux ($A v = \text{constant}$).

मुख्य बिंदु 3

Bernoulli's Law: Invariant sum of pressure, kinetic, and potential fluid energies.

मुख्य बिंदु 4

Terminal Velocity: Steady-state terminal fall speed when viscous drag balances weight.

मुख्य बिंदु 5

Capillary Action: Meniscus surface tension lifting liquid columns through narrow pores.

स्व-मूल्यांकन अभ्यास (Check Your Understanding)

मूल वैचारिक स्पष्टता की जांच के लिए नैदानिक प्रश्न। पहले स्वयं हल करें, फिर उत्तर देखें।

State and prove Bernoulli's Principle.

उत्तर एवं व्याख्या देखें

उत्तर: For the streamline flow of an ideal, incompressible, non-viscous fluid, the sum of pressure energy, kinetic energy, and potential energy per unit volume remains constant throughout the flow: $P + \frac{1}{2}\rho v^2 + \rho g h = \text{constant}$. Follows from Work-Energy theorem applied to fluid element.
P + 1/2 ρv^2 + ρgh = constant.

Explain how an airplane gets dynamic aerodynamic lift using Bernoulli's principle.

उत्तर एवं व्याख्या देखें

उत्तर: An aircraft wing is shaped as an aerofoil: curved on top and flat below. Air rushes faster over the curved upper surface ($v_{\text{top}} > v_{\text{bottom}}$). By Bernoulli's principle, pressure on top is significantly lower ($P_{\text{top}} < P_{\text{bottom}}$), generating an upward dynamic lift force.
Faster air on curved top creates low pressure, generating lift.

A hydraulic car lift has pistons of diameter 5 cm and 30 cm. What force must be applied to the smaller piston to lift a car of mass 1500 kg?

उत्तर एवं व्याख्या देखें

उत्तर: $A_1 = \pi (2.5)^2, A_2 = \pi (15)^2$. $F_2 = mg = 1500 \times 9.8 = 14,700\text{ N}$. By Pascal's law: $F_1 = F_2 \left(\frac{A_1}{A_2}\right) = 14700 \left(\frac{2.5}{15}\right)^2 = 14700 \left(\frac{1}{6}\right)^2 = \frac{14700}{36} \approx 408.3\text{ Newtons}$.
F1 ≈ 408.3 N.

Why do raindrops fall with a constant terminal velocity rather than accelerating continuously under gravity?

उत्तर एवं व्याख्या देखें

उत्तर: As a raindrop falls, upward viscous drag force ($F_v = 6\pi\eta r v$) and buoyancy increase with speed. When the net upward forces balance the downward weight of the drop, net force becomes zero ($a = 0$), and the drop falls at constant terminal velocity.
Viscous drag balances gravity, zeroing acceleration.

What is the excess pressure inside: (i) a liquid drop, (ii) a soap bubble?

उत्तर एवं व्याख्या देखें

उत्तर: (i) Liquid drop (one surface): $\Delta P = \frac{2S}{R}$. (ii) Soap bubble (two surfaces: inner and outer): $\Delta P = \frac{4S}{R}$.
(i) 2S/R, (ii) 4S/R.

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