Consider a completely full cylindrical water tank of height 1.6~m and cross-sectional area 0.5~m2 . It has a small hole…

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

Related Formula

Core Logic

Step 1: Applying Bernoulli's Equation

Step 2: Substitution and Calculation

Pattern Recognition

Chapter Mix

Reference Study Guides

More Fluid Mechanics 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 ( $n$ $n$)