NCERT Solutions for Class 11 Physics Chapter 9 – Mechanical Properties of Fluids

Mechanical Properties of Fluids is a conceptual and scoring chapter in Class 11 Physics that explains the behavior of liquids and gases at rest and in motion. This chapter covers key topics such as pressure in fluids, Pascal’s law, buoyancy, Archimedes’ principle, surface tension, viscosity, Bernoulli’s theorem, and fluid flow, which are important for school exams and competitive exams like JEE and NEET.

NCERT Solutions for Class 11 Physics Chapter 9 – Mechanical Properties of Fluids are prepared strictly according to the latest CBSE syllabus and exam pattern. The solutions are written in simple, step-by-step language with clear derivations, diagrams, and solved numericals, helping students understand fluid mechanics concepts clearly and score well in Class 11 examinations.

NCERT Solutions for Class 11 Physics Chapter 9 – Mechanical Properties of Fluids

NCERT Solutions for Class 11 Physics Chapter 9 – Mechanical Properties of Fluids

Q.1) Explain why:

(a) The blood pressure in humans is greater at the feet than at the brain.

Answer:
The pressure of a liquid is given by the relation

P=hρgP = h\rho g

where ρ\rho is the density of the liquid and hh is the height of the liquid column.

From this relation, it is clear that pressure is directly proportional to height. Since the feet are at a greater vertical distance below the heart compared to the brain, the height of the blood column is greater at the feet. Therefore, the blood pressure at the feet is higher than at the brain.


(b) Atmospheric pressure at a height of about 6 km decreases to nearly half of its value at sea level, although the height of the atmosphere is more than 100 km.

Answer:
The density of air is maximum near sea level and decreases rapidly with increase in height. Most of the atmospheric air is concentrated in the lower layers of the atmosphere.

At a height of about 6 km, the air density becomes nearly half of its value at sea level. Hence, the atmospheric pressure at this height also reduces to nearly half, even though the atmosphere extends beyond 100 km.


(c) Hydrostatic pressure is a scalar quantity even though pressure is force divided by area.

Answer:
When pressure is applied to a liquid, it is transmitted equally in all directions. Hydrostatic pressure does not have any specific direction associated with it.

Since pressure at a point acts equally in all directions, it has only magnitude and no direction. Therefore, hydrostatic pressure is a scalar quantity.


Q.2) Fill in the blanks:

Statement Correct Answer
(a) Surface tension of liquids generally ______ with temperature. Decreases
(b) Viscosity of gases ______ with temperature, whereas viscosity of liquids ______ with temperature. Increases; Decreases
(c) For solids with elastic modulus of rigidity, the shearing force is proportional to ______, while for fluids it is proportional to ______. Shear strain; Rate of shear strain
(d) For a fluid in steady flow, the increase in flow speed at a constriction follows ______. Conservation of mass (and Bernoulli’s principle)
(e) For the model of a plane in a wind tunnel, turbulence occurs at a ______ speed compared to the actual plane. Greater

Q.3) Numerical Problem: Off-shore Structure Suitability

Question:

A vertical off-shore structure is designed to withstand a maximum stress of 10910^9 Pa. Is it suitable for an oil well in the ocean at a depth of 3 km?

Solution:

Maximum stress the structure can withstand

=109Pa= 10^9 \, \text{Pa}

Depth of ocean

d=3×103md = 3 \times 10^3 \, \text{m}

Density of seawater

ρ=103kg m3\rho = 10^3 \, \text{kg m}^{-3}

Acceleration due to gravity

g=9.8m s2g = 9.8 \, \text{m s}^{-2}

Hydrostatic pressure:

P=ρdgP = \rho d g P=103×3×103×9.8P = 10^3 \times 3 \times 10^3 \times 9.8 P=2.94×107PaP = 2.94 \times 10^7 \, \text{Pa}

Conclusion:

Since the pressure exerted by the ocean water (2.94×1072.94 \times 10^7 Pa) is much less than the maximum stress the structure can withstand (10910^9 Pa), the structure is suitable for the oil well.


Q.4) Can Bernoulli’s equation be used for a rapid in a river?

Answer:
No. Bernoulli’s equation is applicable only for streamline (laminar) flow. River rapids involve turbulent flow, so Bernoulli’s equation cannot be applied.


Q.5) Does it matter if one uses gauge pressure instead of absolute pressure in Bernoulli’s equation?

Answer:
No, it does not matter. When Bernoulli’s equation is applied between two points, the atmospheric pressure term cancels out. Hence, either gauge pressure or absolute pressure can be used.


Q.6) Figure Analysis: Venturimeter Flow

Answer:
According to the equation of continuity (Av=constant)(A v = \text{constant}), when the cross-sectional area of the pipe decreases, the speed of the fluid increases.

By Bernoulli’s principle, an increase in speed leads to a decrease in pressure. Since the height of the liquid column in the pipe is directly related to pressure, the water level in pipe 2 should be lower.

Therefore, figure (b) is correct, and figure (a) is incorrect.


FAQs: Class 11 Physics Chapter 9 – Mechanical Properties of Fluids

Q1. Is Mechanical Properties of Fluids important for exams?
Yes, it is a high-weightage chapter in mechanics for Class 11.

Q2. Which topics are most important in this chapter?
Pascal’s law, Archimedes’ principle, Bernoulli’s theorem, surface tension, and viscosity.

Q3. Are numericals asked from this chapter?
Yes, pressure, buoyancy, and Bernoulli-based numericals are very common.

Q4. Are derivations important here?
Yes, derivations related to Bernoulli’s theorem and excess pressure formulas are frequently asked.

Q5. How do NCERT Solutions help?
They provide NCERT-aligned, exam-ready explanations with solved numericals and diagrams.