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

Q3 jee_main_2025_24_jan_morning Excess Pressure and Surface Tension

An air bubble of radius 0.1 cm lies at a depth of 20 cm below the free surface of a liquid of density

1000 kg/m³

If the pressure inside the bubble is 2100 N/m² greater than the atmospheric pressure, then the surface tension of the liquid in SI unit is (use

g=10 m/s²)

A. 0.02
B. 0.1
C. 0.25
D. 0.04

Solution

Related Formula

The absolute pressure inside an air bubble submerged in a liquid is given by :

Pᵢₙ = P₀ + ρ gh + (2T)/(R)

where $P₀$ $P_0$ is atmospheric pressure, $ρ$ $\rho$ is the liquid density, $g$ $g$ is gravity, $h$ $h$ is depth, $T$ $T$ is surface tension, and $R$ $R$ is the radius.

Core Logic

We are given that the difference between the inside pressure and atmospheric pressure is $2100 N/m²$ $2100\text{ N/m}^2$:

Pᵢₙ - P₀ = ρ gh + (2T)/(R) = 2100

Step 1: Numerical Evaluation

Substitute the given values into the relation :

ρ gh = 1000 \times 10 \times 0.20 = 2000 N/m²

Now find the excess pressure from surface tension:

(2T)/(R) = 2100 - 2000 = 100 N/m²

T = (100 \times R)/(2) = 50 \times (0.1 \times 10⁻²) = 0.05 N/m

Pattern Recognition

Total inside pressure accounts for both the hydrostatic pressure of the fluid column ( $ρ gh$ $\rho gh$) and the spherical geometry boundary constraint pressure ( $(2T)/(R)$ $\frac{2T}{R}$).

Chapter Mix

Class 11 Physics: Mechanical Properties of Fluids

Q18 jee_main_2025_28_jan_evening Archimedes Principle

A

400g

solid cube having an edge of length

10cm

floats in water. How much volume of the cube is outside the water? (Given: density of water

= 1000kgm⁻³

)

A. $1400cm³$ $1400\mathrm{cm}^3$
B. $4000cm³$ $4000\mathrm{cm}^3$
C. $400cm³$ $400\mathrm{cm}^3$
D. $600~cm³$ $600~\mathrm{cm}^3$

Solution

Related Formula

By the law of flotation, the weight of a floating body must exactly balance the buoyant force exerted by the displaced fluid volume:

M \cdot g = ρfluid \cdot Vsubmerged \cdot g

Core Logic

Given parameters:

Mass of the cube, $M = 400 g = 0.4 kg$ $M = 400 \text{ g} = 0.4 \text{ kg}$
Total volume of the cube, $Vtotal = (10 cm)³ = 1000 cm³ = 10⁻³ m³$
Density of water, $ρwater = 1000 kg/m³$

Equating weight to buoyant force to find the submerged volume $Vd$ $V_d$ :

0.4 = 1000 \times Vsubmerged

Vsubmerged = (0.4)/(1000) = 4 \times 10⁻⁴ m³ = 400 cm³

Calculate the volume remaining outside the water surface :

Voutside = Vtotal - Vsubmerged

Voutside = 1000 cm³ - 400 cm³ = 600 cm³

Pattern Recognition

The fraction of a floating body's volume that is submerged equals the ratio of the body's density to the fluid's density: $VsubmergedVtotal = ρbodyρfluid$ . Here, the cube's effective density is $0.4 g/cm³$ $0.4 \text{ g/cm}^3$, meaning $40%$ $40\%$ is submerged and $60%$ $60\%$ stays outside.

Chapter Mix

Class 11 Physics: Mechanical Properties of Fluids

Q jee_main_2025_29_jan_morning Pascal\'s Law

In a hydraulic lift, the surface area of the input piston is

6cm²

and that of the output piston is

1500cm²

. If

100N

force is applied to the input piston to raise the output piston by

20cm

, then the work done is ________ kJ.

Numerical Answer. Answer: 5 to 5

Solution

Related Formula

W = F₁ \cdot s₁ = F₂ \cdot s₂

Core Logic

By conservation of liquid volume displacement during output piston elevation :

A₁ \cdot s₁ = A₂ \cdot s₂ 6 \cdot s₁ = 1500 \cdot 20 s₁ = 5000 cm = 50 m

Work performed on input boundary matches :

W = F₁ \cdot s₁ = 100 N \cdot 50 m = 5000 J = 5 kJ

Alternatively via output force profile calculation:

F₂ = F₁ (A₂)/(A₁) = 100 (1500)/(6) = 25000 N

W = F₂ \cdot s₂ = 25000 \cdot 0.2 = 5000 J = 5 kJ

Chapter Mix

Class 11 Physics: Mechanical Properties of Fluids

Q58 jee_main_2024_01_february_morning Bernoulli's Principle

A plane is in level flight at constant speed and each of its two wings has an area of

40~m²

. If the speed of the air is

180~km/h

over the lower wing surface and

252~km/h

over the upper wing surface, the mass of the plane is _______

kg

. (Take air density to be

1~kg m⁻³

and

g = 10~ms⁻²

)

Numerical Answer. Answer: 9600 to 9600

Solution

Related Formula

Bernoulli's pressure balance equation for aerofoils:

Δ P = P₁ - P₂ = (1)/(2)ρ(v₂² - v₁²)

Dynamic Lift force balancing plane weight:

Flift = Δ P \cdot Atotal = mg

Core Logic

Convert velocity limits to SI units:

v₁ = 180~km/h = 180 \times (5)/(18) = 50~ms⁻¹

v₂ = 252~km/h = 252 \times (5)/(18) = 70~ms⁻¹

Total effective wing area layout (2 wings):

Atotal = 2 \times 40 = 80~m²

Step 1: Calculate Mass Balance

Substitute these values into the dynamic lift equation:

mg = (1)/(2) ρ (v₂² - v₁²) Atotal

m(10) = (1)/(2) \times 1 \times (70² - 50²) \times 80

10m = 40 \times (4900 - 2500) = 40 \times 2400 = 96000

m = 9600~kg

Pattern Recognition

Remember to multiply individual wing areas by 2 for standard multi-wing lift structures ( $Atotal = 2A$ $A_{\text{total}} = 2A$).

Chapter Mix

Class 11 Physics: Mechanical Properties of Fluids

Q jee_main_2024_29_january_evening Surface Tension

A small liquid drop of radius

R

is divided into

27

identical liquid drops. If the surface tension is

T

, then the work done in the process will be:

A. $8π R² T$ $8\pi R^2 T$
B. $3π R² T$ $3\pi R^2 T$
C. $(1)/(8)π R² T$ $\frac{1}{8}\pi R^2 T$
D. $4π R² T$ $4\pi R^2 T$

Solution

Related Formula

The work done in changing the surface area of a liquid is given by:

W = T \cdot Δ A

where:

$T$ $T$ is the surface tension of the liquid.
$Δ A = Af - Aᵢ$ $\Delta A = A_f - A_i$ is the change in the total surface area.

Core Logic

Since the total volume remains constant during splitting:

$Vᵢ = Vf$ $V_i = V_f$

(4)/(3)π R³ = 27 \times (4)/(3)π r³

R³ = 27r³ r = (R)/(3)

Step 1: Calculate the Change in Surface Area

Initial surface area of the single drop:

$Aᵢ = 4π R²$ $A_i = 4\pi R^2$

Final surface area of $27$ $27$ small drops:

Af = 27 \times (4π r²) = 27 \times 4π ((R)/(3))²

Af = 27 \times 4π (R²)/(9) = 12π R²

Change in surface area:

Δ A = Af - Aᵢ = 12π R² - 4π R² = 8π R²

Step 2: Calculate Work Done

Substituting the change in area into the work done formula:

W = T \cdot Δ A = 8π R² T

Pattern Recognition

For splitting a large drop of radius $R$ $R$ into $n$ $n$ identical small drops, the change in surface area is given by $Δ A = 4π R² (n1/3 - 1)$ $\Delta A = 4\pi R^2 (n^{1/3} - 1)$. Substituting $n = 27$ $n = 27$ gives $Δ A = 4π R² (3 - 1) = 8π R²$ , leading immediately to $8π R² T$ $8\pi R^2 T$.

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 5

Solution

Related Formula

Core Logic

Step 1: Numerical Evaluation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Mass Balance

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate the Change in Surface Area

Step 2: Calculate Work Done

Pattern Recognition

Chapter Mix

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