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Mechanical Properties of Fluids appeared 31 times across 3 years — 3.6% of Physics. This question is from Viscosity and Terminal Velocity.

Year 2026 2025 2024 Total
Questions 9 14 8 31

In the experiment for measurement of viscosity η of given liquid with a ball having radius R , consider following statements. A. Graph between terminal velocity V and R will be a parabola B. The terminal velocities of different diameter balls are constant for a given liquid. C. Measurement of terminal velocity is dependent on the temperature. D. This experiment can be utilized to assess the density of a given liquid. E. If balls are dropped with some initial speed, the value of η will change. Choose the correct answer from the options given below:

Solution & Explanation

Related Formula
VT = (2)/(9) R² gη (d - ρ)
Core Logic

Let us check each statement sequentially:

Statement A: Since VT ∝ R², the graph between terminal velocity V and radius R is a quadratic curve, i.e., a parabola. (True)

Statement B: Different diameters imply different radii, hence terminal velocity will change. It is not constant. (False)

Statement C: Viscosity of a liquid is heavily temperature-dependent, hence VT changes with temperature. (True)

Statement D: Knowing the parameters allows calculation of liquid density ρ. (True)

Statement E: η is a material property of the fluid and does not change based on the initial dropping speed of the ball. (False)

Step 1: Final Conclusion

Statements A, C, and D are correct, which matches option (2).

Pattern Recognition

Terminal velocity problems rely entirely on parsing the proportional relations hidden in Stokes' Law formulation. Note that viscosity coefficient is an intrinsic characteristic parameter independent of kinematics.

Chapter Mix

Class 11 Physics: Mechanical Properties of Fluids

Reference Study Guides

More Mechanical Properties of Fluids Previous-Year Questions — Page 6

Q32 jee_main_2024_27_jan_morning Viscosity and Surface Tension
Given below are two statements: Statement (I): Viscosity of gases is greater than that of liquids. Statement (II): Surface tension of a liquid decreases due to the presence of insoluble impurities. In the light of the above statements, choose the most appropriate answer from the options given below:
  • A. Statement I is correct but Statement II is incorrect
  • B. Statement I is incorrect but Statement II is correct
  • C. Both Statement I and Statement II are incorrect
  • D. Both Statement I and Statement II are correct

Solution

Core Logic

Statement (I): Liquids have much stronger intermolecular forces compared to gases, leading to significantly higher viscosity in liquids than in gases. Thus, Statement I is incorrect.

Statement (II): The presence of insoluble impurities (like soap or detergents) disrupts the cohesive forces between liquid molecules at the surface, which decreases the surface tension. Thus, Statement II is correct.

Pattern Recognition

Viscosity in liquids decreases with temperature, whereas in gases it increases with temperature due to molecular collisions. Insoluble impurities act as surface-active agents that lower surface tension cohesive stability.

Chapter Mix

Class 11 Physics: Mechanical Properties of Fluids

Q33 jee_main_2024_29_jan_morning Surface Tension and Capillarity
Given below are two statements: Statement I: If a capillary tube is immersed first in cold water and then in hot water, the height of capillary rise will be smaller in hot water. Statement II: If a capillary tube is immersed first in cold water and then in hot water, the height of capillary rise will be smaller in cold water. In the light of the above statements, choose the most appropriate from the options given below:
  • A. Both Statement I and Statement II are true
  • B. Both Statement I and Statement II are false
  • C. Statement I is true but Statement II is false
  • D. Statement I is false but Statement II is true

Solution

Related Formula

The height of capillary rise (h) is given by:

h = (2T θ)/(ρ g r)

where, T = surface tension of the liquid θ = angle of contact ρ = density of the liquid r = radius of the capillary tube

Core Logic

As the temperature of water increases, its intermolecular cohesive forces decrease. This leads to a decrease in surface tension (T).

Since h ∝ T (assuming ρ and θ remain relatively constant), height of capillary rise decreases with increase in temperature.

Step 1: Evaluate Statement I and II

Because Thot lt Tcold, it follows that hhot lt hcold.

Pattern Recognition

Remember the key physical dependence: Temperature Up Surface Tension Down Capillary Rise Down. This basic trend of cohesive/adhesive properties versus thermal energy frequently appears in competitive conceptual physical chemistry/fluid physics questions.

Chapter Mix

Class 11 Physics: Mechanical Properties of Fluids

Q60 jee_main_2024_29_jan_morning Bernoulli's Principle
In a test experiment on a model aeroplane in wind tunnel, the flow speeds on the upper and lower surfaces of the wings are 70~ms⁻¹ and 65~ms⁻¹ respectively. If the wing area is 2~m² the lift of the wing is ________ N. (Given density of air = 1.2~kg m⁻³)
Numerical Answer. Answer: 810 to 810

Solution

Related Formula

From Bernoulli's Principle (ignoring small height differences across the wing thickness):

P₁ + (1)/(2) ρ v₁² = P₂ + (1)/(2) ρ v₂² Δ P = P₂ - P₁ = (1)/(2) ρ (v₁² - v₂²)

Net aerodynamic lift force (F) acting on wing area A is:

F = Δ P · A = (1)/(2) ρ (v₁² - v₂²) A
Core Logic

Given values:

  • Flow speed on upper surface (v₁) = 70 ~ms⁻¹
  • Flow speed on lower surface (v₂) = 65 ~ms⁻¹
  • Wing area (A) = 2 ~m²
  • Density of air (ρ) = 1.2 ~kg· m⁻³
Step 1: Compute Lift Force

Substituting these metrics directly into the dynamic lift equation:

F = (1)/(2) × 1.2 × (70² - 65²) × 2

Notice that the factors (1)/(2) and 2 cancel out perfectly:

F = 1.2 × (4900 - 4225) F = 1.2 × 675 = 810 ~N

Therefore, the net lift force on the wing is 810 ~N.

Pattern Recognition

Use the algebraic difference of squares shortcut a² - b² = (a-b)(a+b) to solve velocity square differences quickly: 70² - 65² = (70-65)(70+65) = 5 × 135 = 675, bypassing squaring heavy multi-digit calculations.

Chapter Mix

Class 11 Physics: Mechanical Properties of Fluids

Q52 jee_main_2024_30_january_evening Surface Energy of Drops
A big drop is formed by coalescing 1000 small identical drops of water. If E₁ be the total surface energy of 1000 small drops of water and E₂ be the surface energy of single big drop of water, the E₁:E₂ is x:1 where x =
Numerical Answer. Answer: 10 to 10

Solution

Related Formula
Vinitial = Vfinal

E = S × A

Core Logic

When small drops coalesce to form a large drop, the total volume is conserved.

1000 × (4)/(3) π r³ = (4)/(3) π R³ R³ = 1000 r³ R = 10r
Step 1: Calculate Surface Energies

The total surface energy of 1000 small drops (E₁) is:

E₁ = 1000 × 4π r² × S

The surface energy of the single big drop (E₂) is:

E₂ = 4π R² × S = 4π (10r)² × S = 100 × 4π r² × S
Step 2: Find the Ratio
E₁E₂ = (1000)/(100) = (10)/(1)

Thus, the ratio is 10:1, which means x = 10.

Pattern Recognition

When N droplets merge to form one big drop, the radius scales as R = N1/3 r. The ratio of total initial surface energy to final surface energy is N1/3.

Chapter Mix

Class 11 Physics: Mechanical Properties of Fluids

Q47 jee_main_2024_31_jan_evening Viscosity and Terminal Velocity
A small spherical ball of radius r, falling through a viscous medium of negligible density has terminal velocity 'v'. Another ball of the same mass but of radius 2r, falling through the same viscous medium will have terminal velocity:
  • A. (v)/(2)
  • B. (v)/(4)
  • C. 4v
  • D. 2v

Solution

Related Formula

At terminal velocity, downward force equals upward drag (assuming negligible buoyancy):

Mg = 6π η r v v = (Mg)/(6π η r)
Core Logic

Since the density of the medium is negligible, we ignore buoyant forces. The mass M of the ball is specified to remain the same in both cases, despite the change in radius (implying the material density of the second ball is lower).

Step 1: Setup Proportionality

Since M, g, and η are all constants:

v ∝ (1)/(r)
Step 2: Evaluating the Ratio

For the second ball, r' = 2r. Therefore, the new terminal velocity v' is:

v' = v × ((r)/(r')) = v × ((r)/(2r)) = (v)/(2)
Pattern Recognition

Read the constraints carefully. Usually, questions keep material density uniform (v ∝ r²). However, this specifically says "same mass". This shifts the formula dependency from v ∝ r² entirely to v ∝ 1/r because M acts as a constant numerator.

Chapter Mix

Class 11 Physics: Mechanical Properties of Fluids

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