Consider following statements: A. Surface tension arises due to extra energy of the molecules at the interior as compared to the molecules at the surface, of a liquid. B. As the temperature of liquid rises, the coefficient of viscosity increases. C. As the temperature of gas increases, the coefficient of viscosity increases. D. The onset of turbulence is determined by Reynold's number. E. In a steady flow two stream lines never intersect. Choose the correct answer from the options given below:

Let's audit each fluid mechanics assertion: Statement A: Incorrect. Surface tension arises because surface molecules possess higher potential energy compared to interior bulk molecules due to net cohesive forces pulling inward. Statement B: Incorrect. With rising temperatures, cohesive forces in liquids weaken, causing viscosity to decrease. Statement C: Correct. In gases, viscosity is governed by molecular collisions. Higher temperature leads to increased thermal activity and momentum exchange, increasing viscosity. Statement D: Correct. Critical velocity and turbulence parameters are completely defined by the dimensionless Reynolds number. Statement E: Correct. If streamlines intersected, a fluid particle at that spatial node would have two distinct velocity directions, violating steady-state constraints. Statements C, D, and E are correct, leading to option (2). Contrast Liquid vs Gas viscosity trends under temperature increments: Liquids down (intermolecular bounds weaken), Gases up (random thermal diffusion and collision rates escalate). Class 11 Physics: Mechanical Properties of Fluids

Consider following statements: A. Surface tension arises due to extra energy of the molecules at the interior as compared to the molecules at the surface, of a liquid. B. As the temperature of liquid

Let's audit each fluid mechanics assertion: Statement A: Incorrect. Surface tension arises because surface molecules possess higher potential energy compared to interior bulk molecules due to net cohesive forces pulling inward. Statement B: Incorrect. With rising temperatures, cohesive forces in liquids weaken, causing viscosity to decrease. Statement C: Correct. In gases, viscosity is governed by molecular collisions. Higher temperature leads to increased thermal activity and momentum exchange,

Consider following statements: A. Surface tension arises due to extra energy of the molecules at the interior as compared to the molecules at the surface, of a liquid. B. As the temperature of liquid rises, the coefficient of viscosity increases. C. As the temperature of gas increases, the coefficient of viscosity increases. D. The onset of turbulence is determined by Reynold's number. E. In a steady flow two stream lines never intersect. Choose the correct answer from the options given below:

Let's audit each fluid mechanics assertion: Statement A: Incorrect. Surface tension arises because surface molecules possess higher potential energy compared to interior bulk molecules due to net cohesive forces pulling inward. Statement B: Incorrect. With rising temperatures, cohesive forces in liquids weaken, causing viscosity to decrease. Statement C: Correct. In gases, viscosity is governed by molecular collisions. Higher temperature leads to increased thermal activity and momentum exchange, increasing viscosity. Statement D: Correct. Critical velocity and turbulence parameters are completely defined by the dimensionless Reynolds number. Statement E: Correct. If streamlines intersected, a fluid particle at that spatial node would have two distinct velocity directions, violating steady-state constraints. Statements C, D, and E are correct, leading to option (2). Contrast Liquid vs Gas viscosity trends under temperature increments: Liquids down (intermolecular bounds weaken), Gases up (random thermal diffusion and collision rates escalate). Class 11 Physics: Mechanical Properties of Fluids

Consider following statements: A. Surface tension arises due to extra energy of the molecules at the interior as compared to the molecules at the surface, of a liquid. B. As the temperature of liquid

Let's audit each fluid mechanics assertion: Statement A: Incorrect. Surface tension arises because surface molecules possess higher potential energy compared to interior bulk molecules due to net cohesive forces pulling inward. Statement B: Incorrect. With rising temperatures, cohesive forces in liquids weaken, causing viscosity to decrease. Statement C: Correct. In gases, viscosity is governed by molecular collisions. Higher temperature leads to increased thermal activity and momentum exchange,

Consider following statements: A. Surface tension arises due to extra energy of the molecules at the interior as…

Year	2026	2025	2024	Total
Questions	9	14	8	31

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 ( $h$ $h$) is given by:

h = (2T θ)/(ρ g r)

where, $T$ $T$ = surface tension of the liquid $θ$ $\theta$ = angle of contact $ρ$ $\rho$ = density of the liquid $r$ $r$ = 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 ( $T$ $T$).

Since $h \propto T$ $h \propto T$ (assuming $ρ$ $\rho$ and $θ$ $\theta$ remain relatively constant), height of capillary rise decreases with increase in temperature.

Step 1: Evaluate Statement I and II

Because $Thot lt Tcold$ $T_{\text{hot}} \lt T_{\text{cold}}$, it follows that $hhot lt hcold$ $h_{\text{hot}} \lt h_{\text{cold}}$.

Statement I states: the height of capillary rise will be smaller in hot water. This is True.
Statement II states: the height of capillary rise will be smaller in cold water. This is False.

Pattern Recognition

Remember the key physical dependence: Temperature Up $\implies$ Surface Tension Down $\implies$ 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 ( $F$ $F$) acting on wing area $A$ $A$ is:

F = Δ P \cdot A = (1)/(2) ρ (v₁² - v₂²) A

Core Logic

Given values:

Flow speed on upper surface ( $v₁$ $v_1$) = $70 ~ms⁻¹$ $70 \mathrm{~ms}^{-1}$
Flow speed on lower surface ( $v₂$ $v_2$) = $65 ~ms⁻¹$ $65 \mathrm{~ms}^{-1}$
Wing area ( $A$ $A$) = $2 ~m²$ $2 \mathrm{~m}^2$
Density of air ( $ρ$ $\rho$) = $1.2 ~kg\cdot m⁻³$ $1.2 \mathrm{~kg\cdot m}^{-3}$

Step 1: Compute Lift Force

Substituting these metrics directly into the dynamic lift equation:

F = (1)/(2) \times 1.2 \times (70² - 65²) \times 2

Notice that the factors $(1)/(2)$ $\frac{1}{2}$ and $2$ $2$ cancel out perfectly:

F = 1.2 \times (4900 - 4225)

F = 1.2 \times 675 = 810 ~N

Therefore, the net lift force on the wing is $810 ~N$ $810 \mathrm{~N}$.

Pattern Recognition

Use the algebraic difference of squares shortcut $a² - b² = (a-b)(a+b)$ $a^2 - b^2 = (a-b)(a+b)$ to solve velocity square differences quickly: $70² - 65² = (70-65)(70+65) = 5 \times 135 = 675$ , 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

$E = S \times A$ $E = S \times A$

Core Logic

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

1000 \times (4)/(3) π r³ = (4)/(3) π R³

R³ = 1000 r³ R = 10r

Step 1: Calculate Surface Energies

The total surface energy of $1000$ $1000$ small drops ( $E₁$ $\mathrm{E}_1$) is:

E₁ = 1000 \times 4π r² \times S

The surface energy of the single big drop ( $E₂$ $\mathrm{E}_2$) is:

E₂ = 4π R² \times S = 4π (10r)² \times S = 100 \times 4π r² \times S

Step 2: Find the Ratio

E₁E₂ = (1000)/(100) = (10)/(1)

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

Pattern Recognition

When $N$ $N$ droplets merge to form one big drop, the radius scales as $R = N1/3 r$ $R = N^{1/3} r$. The ratio of total initial surface energy to final surface energy is $N1/3$ $N^{1/3}$.

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. $(v)/(2)$ $\frac{v}{2}$
B. $(v)/(4)$ $\frac{v}{4}$
C. $4v$ $4v$
D. $2v$ $2v$

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 $M$ $M$ 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 $M$ $M$, $g$ $g$, and $η$ $\eta$ are all constants:

v \propto (1)/(r)

Step 2: Evaluating the Ratio

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

v' = v \times ((r)/(r')) = v \times ((r)/(2r)) = (v)/(2)

Pattern Recognition

Read the constraints carefully. Usually, questions keep material density uniform ( $v \propto r²$ $v \propto r^2$). However, this specifically says "same mass". This shifts the formula dependency from $v \propto r²$ $v \propto r^2$ entirely to $v \propto 1/r$ $v \propto 1/r$ because $M$ $M$ acts as a constant numerator.

Chapter Mix

Class 11 Physics: Mechanical Properties of Fluids

Solution & Explanation

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

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

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Evaluate Statement I and II

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute Lift Force

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Surface Energies

Step 2: Find the Ratio

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Setup Proportionality

Step 2: Evaluating the Ratio

Pattern Recognition

Chapter Mix

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