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:

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. 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. Class 11 Physics: Mechanical Properties of Fluids

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

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. Viscosity in liquids decreases with temperature, whereas in gases it increases with temperature due

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:

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. 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. Class 11 Physics: Mechanical Properties of Fluids

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

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. Viscosity in liquids decreases with temperature, whereas in gases it increases with temperature due

Given below are two statements: Statement (I): Viscosity of gases is greater than that of liquids. Statement (II)…

Year	2026	2025	2024	Total
Questions	9	15	8	32

Q21 jee_main_2025_02_april_morning Hydrostatic Pressure

A vessel with square cross-section and height of

6~m

is vertically partitioned. A small window of

100~cm²

with hinged door is fitted at a depth of

3~m

in the partition wall. One part of the vessel is filled completely with water and the other side is filled with the liquid having density

1.5\times 10³~kg/m³

. What force one needs to apply on the hinged door so that it does not get opened? (Acceleration due to gravity

= 10~m/s²

)

Numerical Answer. Answer: 150 to 150

Solution

Related Formula

P = P₀ + ρ g h

F = Δ P \cdot A

Core Logic

Let's analyze the pressure acting on the window at depth $h = 3~m$ $h = 3\mathrm{~m}$ from both sides:

Water side (density $ρw = 1.0 \times 10³~kg/m³$ ):

Pw = P₀ + ρw g h

Denser liquid side (density $ρl = 1.5 \times 10³~kg/m³$ ):

Pl = P₀ + ρl g h

The net pressure difference pushing on the window is:

Δ P = Pl - Pw = (ρl - ρw) g h

Δ P = (1.5 \times 10³ - 1.0 \times 10³) \times 10 \times 3 = 500 \times 30 = 15000~N/m²

The window has area $A = 100~cm² = 100 \times 10⁻⁴~m² = 10⁻²~m²$ .

To keep the door from opening, we must apply a force balancing the pressure difference:

F = Δ P \cdot A = 15000 \times 10⁻² = 150~N

Step 1: Final Conclusion The required force is$ Step 1: Final Conclusion The required force is $150\mathrm{~N}. Pattern Recognition For a vertical interface between two fluids, the net pressure difference is simply given by$. Pattern Recognition For a vertical interface between two fluids, the net pressure difference is simply given by $\Delta P = \Delta \rho \cdot g h. The ambient atmospheric pressure$. The ambient atmospheric pressure $P_0$ cancels out since it acts on both sides of the window. Chapter Mix Class 11 Physics: Mechanical Properties of Fluids

Q jee_main_2025_03_april_evening Work Done by Gravity in Connecting Vessels

Two cylindrical vessels of equal cross sectional area of

2~m²

contain water upto height

10~m

and

6~m

, respectively. If the vessels are connected at their bottom then the work done by the force of gravity is : (Density of water is

10³~kg/m³

and

g=10~m/s²

)

A. $1 \times 10⁵~J$ $1 \times 10^{5}\mathrm{~J}$
B. $4 \times 10⁴~J$ $4 \times 10^{4}\mathrm{~J}$
C. $6 \times 10⁴~J$ $6 \times 10^{4}\mathrm{~J}$
D. $8 \times 10⁴~J$ $8 \times 10^{4}\mathrm{~J}$

Solution

Related Formula

The gravitational potential energy $U$ $U$ of a liquid column of mass $m$ $m$ and height $h$ $h$ is evaluated relative to its bottom by placing its total mass at its center of mass ( $h/2$ $h/2$):

U = m g ((h)/(2)) = (ρ A h) g ((h)/(2)) = (1)/(2) ρ A g h²

Work done by the force of gravity ( $W$ $W$) equals the negative change in potential energy:

W = -Δ U = Uᵢ - Uf

Core Logic

Since the vessels are identical and connected at the bottom, water flows from the higher column to the lower one until their final heights equalize at:

hf = (10 + 6)/(2) = 8~m

Step 1: Calculate Initial Potential Energy ( $Uᵢ$ $U_i$)

Let the reference level $U=0$ $U=0$ be at the bottom:

Uᵢ = U₁ + U₂ = (1)/(2) ρ A g h₁² + (1)/(2) ρ A g h₂²

Uᵢ = (1)/(2) ρ A g (10² + 6²) = (1)/(2) ρ A g (100 + 36) = 68 ρ A g

Work Done by Gravity in Connecting Vessels

Step 2: Calculate Final Potential Energy ( $Uf$ $U_f$)

Both vessels equalize to $hf = 8~m$ $h_f = 8\mathrm{~m}$:

Uf = 2 \times [ (1)/(2) ρ A g hf² ] = ρ A g (8²) = 64 ρ A g

Step 3: Work Done by Gravity ( $W$ $W$)

W = Uᵢ - Uf = 68 ρ A g - 64 ρ A g = 4 ρ A g

Substitute the given values ( $ho = 10³~kg/m³$ $ ho = 10^3\mathrm{~kg/m}^3$, $A = 2~m²$ $A = 2\mathrm{~m}^2$, $g = 10~m/s²$ $g = 10\mathrm{~m/s}^2$):

W = 4 \times 10³ \times 2 \times 10 = 8 \times 10⁴~J

Pattern Recognition For leveling liquids between two identical connected columns, the shift in center of mass always simplifies. The loss in potential energy is given by$ Pattern Recognition For leveling liquids between two identical connected columns, the shift in center of mass always simplifies. The loss in potential energy is given by $\Delta U = \frac{1}{4} \rho A g (h_1 - h_2)^2. Applying this directly:

Δ U = (1)/(4) \times 10³ \times 2 \times 10 \times (10 - 6)² = 5000 \times 16 = 8 \times 10⁴~J

$

Chapter Mix

Class 11 Physics: Mechanical Properties of Fluids

Q12 jee_main_2025_03_april_evening Viscosity and Terminal Velocity

A solid steel ball of diameter

3.6~mm

acquired terminal velocity

2.45\times10⁻²~m/s

while falling under gravity through an oil of density

925~kg~m⁻³

. Take density of steel as

7825~kg~m⁻³

and g as

9.8~m/s²

. The viscosity of the oil in SI unit is :

A. 2.18
B. 2.38
C. 1.68
D. 1.99

Solution

Related Formula

Terminal velocity $vₜ$ $v_t$ of a spherical body falling through a viscous fluid is given by Stokes' Law:

vₜ = (2)/(9) r² g (ρbody - ρfluid)η

Hence, the viscosity coefficient $η$ $\eta$ is:

η = (2)/(9) r² g (ρbody - ρfluid)vₜ

Core Logic

Given parameters:

Diameter $d = 3.6~mm \Rightarrow$ $d = 3.6\mathrm{~mm} \Rightarrow$ radius $r = 1.8~mm = 1.8 \times 10⁻³~m$
Terminal velocity $vₜ = 2.45 \times 10⁻²~m/s$
Liquid density $ρfluid = 925~kg/m³$
Steel density $ρbody = 7825~kg/m³$
Acceleration due to gravity $g = 9.8~m/s²$ $g = 9.8\mathrm{~m/s}^2$

Step 1: Substitute parameters into formula

η = (2)/(9) \times (1.8 \times 10⁻³)² \times 9.8 \times (7825 - 925)2.45 \times 10⁻²

η = (2)/(9) \times 3.24 \times 10⁻⁶ \times 9.8 \times 69002.45 \times 10⁻²

Step 2: Solve the numeric calculation

η = 2 \times 0.36 \times 10⁻⁶ \times 9.8 \times 69002.45 \times 10⁻²

η = 0.72 \times 9.8 \times 6900 \times 10⁻⁶2.45 \times 10⁻²

η = 48700.8 \times 10⁻⁶2.45 \times 10⁻² = (0.0487008)/(0.0245) \approx 1.99~Pa\cdot s

Pattern Recognition Be careful when converting diameter to radius ($ Pattern Recognition Be careful when converting diameter to radius ($r = d/2$). Always ensure all numerical parameters are converted cleanly to SI base units (meters, kilograms, seconds) before applying Stokes' formula. Chapter Mix Class 11 Physics: Mechanical Properties of Fluids

Q22 jee_main_2025_03_april_evening Excess Pressure in Soap Bubbles

The excess pressure inside a soap bubble A in air is half the excess pressure inside another soap bubble B in air. If the volume of the bubble A is

n

times the volume of the bubble B, then the value of

n

is ________.

Numerical Answer. Answer: 8 to 8

Solution

Related Formula

Excess pressure inside a soap bubble in air (which has two liquid-gas interfaces) is given by:

Δ P = (4T)/(R) \Rightarrow R \propto (1)/(Δ P)

The volume $V$ $V$ of a spherical bubble of radius $R$ $R$ is:

V = (4)/(3)π R³ \Rightarrow V \propto R³ \propto ((1)/(Δ P))³

Core Logic

Given state:

Δ PA = (1)/(2) Δ PB \Rightarrow (Δ PB)/(Δ PA) = 2

Step 1: Calculate the ratio of radii

(RA)/(RB) = (Δ PB)/(Δ PA) = 2

Step 2: Calculate volume scaling parameter ( $n$ $n$)

VA = n VB \Rightarrow n = (VA)/(VB) = ((RA)/(RB))³ = (2)³ = 8

Pattern Recognition

Volume scale factors depend on the cube of linear scale factors ( $V \propto R³$ $V \propto R^3$). Since radius is inversely proportional to excess pressure, a halving of excess pressure leads to doubling of radius, scaling volume by $2³ = 8$ $2^3 = 8$.

Chapter Mix

Class 11 Physics: Mechanical Properties of Fluids

Q22 jee_main_2025_08_april_evening Bulk Modulus

A sample of a liquid is kept at

1~atm

. It is compressed to

5~atm

which leads to change of volume of

0.8~cm³

. If the bulk modulus of the liquid is

2~GPa

, the initial volume of the liquid was ________ litre. (Take

1~atm = 10⁵~Pa

)

Numerical Answer. Answer: 4 to 4

Solution

Related Formula

B = -(Δ P)/(Δ V / V) V = B (-Δ V)/(Δ P)

where, $B$ $B$ = Bulk modulus of liquid $Δ P$ $\Delta P$ = change in pressure $Δ V$ $\Delta V$ = change in volume $V$ $V$ = initial volume

Core Logic

Given parameters:

Initial pressure, $Pᵢ = 1~atm$ $P_i = 1\mathrm{~atm}$
Final pressure, $Pf = 5~atm$ $P_f = 5\mathrm{~atm}$
Change in pressure, $Δ P = Pf - Pᵢ = 4~atm = 4 \times 10⁵~Pa$
Change in volume, $Δ V = -0.8~cm³ = -0.8 \times 10⁻⁶~m³$
Bulk modulus, $B = 2~GPa = 2 \times 10⁹~Pa$

Step 1: Compute Initial Volume

Substitute the parameters into the formula:

V = 2 \times 10⁹ \times 0.8 \times 10⁻⁶4 \times 10⁵

V = (1.6 \times 10³)/(4 \times 10⁵) = 0.4 \times 10⁻² = 4 \times 10⁻³~m³

Convert cubic meters to litres:

1~m³ = 1000~litres

V = 4 \times 10⁻³ \times 1000 = 4~litres

Thus, the initial volume was $4~litres$ $4\mathrm{~litres}$.

Pattern Recognition

Sees: Bulk modulus definition calculation. Trap: Watch out for unit conversions: $1~GPa = 10⁹~Pa$ $1\mathrm{~GPa} = 10^9\mathrm{~Pa}$, and $1~cm³ = 10⁻⁶~m³$ . Finally, express the answer in Litres, where $1~L = 10⁻³~m³$ $1\mathrm{~L} = 10^{-3}\mathrm{~m}^3$. ✓

Chapter Mix

Class 11 Physics: Mechanical Properties of Fluids

Solution & Explanation

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Mechanical Properties of Fluids Previous-Year Questions — Page 3

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Step 1: Final Conclusion

Pattern Recognition

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Initial Potential Energy (Uᵢ$U_i$)

Step 2: Calculate Final Potential Energy (Uf$U_f$)

Step 3: Work Done by Gravity (W$W$)

Pattern Recognition

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Substitute parameters into formula

Step 2: Solve the numeric calculation

Pattern Recognition

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate the ratio of radii

Step 2: Calculate volume scaling parameter (n$n$)

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute Initial Volume

Pattern Recognition

Chapter Mix

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Step 1: Calculate Initial Potential Energy ( $Uᵢ$ $U_i$)

Step 2: Calculate Final Potential Energy ( $Uf$ $U_f$)

Step 3: Work Done by Gravity ( $W$ $W$)

Step 2: Calculate volume scaling parameter ( $n$ $n$)