A capillary tube of radius 0.1 mm is partly dipped in water (surface tension 70~dyn/cm and glass water contact angle…

Year	2026	2025	2024	Total
Questions	9	15	8	32

Q45 jee_main_2026_23_january_evening Viscosity and Terminal Velocity

A small metallic sphere of diameter 2 mm and density

10.5 g/cm³

is dropped in glycerine having viscosity 10 Poise and density

1.5 g/cm³

respectively. The terminal velocity attained by the sphere is ____ cm/s. (

π = (22)/(7) and g = 10 m/s²

)

A. $2.0$ $2.0$
B. $1.0$ $1.0$
C. $3.0$ $3.0$
D. $1.5$ $1.5$

Solution

Related Formula

VT = (2r² g)/(9η) (ρb - ρ)

Core Logic

Given data: Radius $r = 1 mm = 0.1 cm$ . Density of sphere $ρb = 10.5 g/cm³$ $\rho_b = 10.5 \, \mathrm{g/cm}^3$. Density of liquid $ρ = 1.5 g/cm³$ $\rho_{\ell} = 1.5 \, \mathrm{g/cm}^3$. Viscosity $η = 10 Poise = 10 g/cm\cdot s$ (since 1 Poise = 1 CGS unit of viscosity). Gravity $g = 10 m/s² = 1000 cm/s²$ (Wait, standard gravity for CGS is $980$ $980$, but the problem specifies $g = 10 m/s² = 1000 cm/s²$ ).

Step 1: Calculation in CGS Units

VT = (2)/(9) \cdot ((0.1)² \times 1000)/((10)) \cdot (10.5 - 1.5)

VT = (2)/(9) \cdot (0.01 \times 1000)/(10) \cdot (9)

VT = 2 \cdot (10)/(10) \cdot 1

VT = 2 cm/s

Pattern Recognition

Watch the units closely. Poise is the CGS unit for viscosity, so converting all values to CGS (cm, g, s) makes the calculation direct and prevents factor-of-10 errors common when mixing SI and CGS.

Chapter Mix

Class 11 Physics: Mechanical Properties of Fluids

Q50 jee_main_2026_24_january_morning Terminal Velocity

Sixty four rain drops of radius 1 mm each falling down with a terminal velocity of 10 cm/s coalesce to form a bigger drop. The terminal velocity of bigger drop is ____ cm/s.

Numerical Answer. Answer: 160 to 160

Solution

Related Formula

VT = (2r² g)/(9η) (ρ - σ)

$VT \propto r²$ $V_T \propto r^2$

Core Logic

Coalescing rain drops volume conservation

When 64 small drops (radius $R₁$ $R_1$) coalesce to form a bigger drop (radius $R₂$ $R_2$), mass and volume are conserved.

64 ((4)/(3) π R₁³) = (4)/(3) π R₂³

R₂³ = 64 R₁³ R₂ = 4 R₁

Step 1: Terminal Velocity Ratio

Since terminal velocity $VT \propto R²$ $V_T \propto R^2$:

((VT)₁)/((VT)₂) = ( (R₁)/(R₂) )² = ( (1)/(4) )² = (1)/(16)

Given $(VT)₁ = 10 cm/s$ $(V_T)_1 = 10 \text{ cm/s}$:

(10)/((VT)₂) = (1)/(16)

(VT)₂ = 160 cm/s

Pattern Recognition

When $N$ $N$ identical drops coalesce, the new radius is $N1/3$ $N^{1/3}$ times the old radius. Consequently, the new terminal velocity scales precisely as $N2/3$ $N^{2/3}$. For $N=64$ $N=64$, $642/3 = 16$ $64^{2/3} = 16$. Just multiply initial $v$ $v$ by 16.

Chapter Mix

Class 11 Physics: Mechanical Properties of Fluids

Q37 jee_main_2026_24_january_evening Buoyancy and Archimedes Principle

A cubical block of density

ρb = 600 kg/m³

floats in a liquid of density

ρₑ = 900 kg/m³

. If the height of block is

H = 8.0 cm

then height of the submerged part is ____ cm.

A. $7.3$ $7.3$
B. $4.3$ $4.3$
C. $6.3$ $6.3$
D. $5.3$ $5.3$

Solution

Related Formula

Fbuoyancy = Mg

ρliquid Vsubmerged g = ρblock Vtotal g

Core Logic

For a floating block, the weight of the block equals the buoyant force exerted by the liquid.

$M g = Fb$ $M g = F_b$

ρb \times A \times H \times g = ρₑ \times A \times h \times g

Step 1: Calculation

600 \times 8 cm = 900 \times h

h = (600 \times 8)/(900)

h = (16)/(3) cm

h \approx 5.33 cm

Pattern Recognition

Submerged fraction is exactly the ratio of densities: $(h)/(H) = ρobjectρfluid$ .

Chapter Mix

Class 11 Physics: Mechanical Properties of Fluids

Q50 jee_main_2026_24_january_evening Surface Tension and Work Done

A soap bubble of surface tension 0.04 N/m is blown to a diameter of 7 cm. If (15000 - x) μJ of work is done in blowing it further to make its diameter 14 cm, then the value of x is ____.

(π = 2 2 / 7)

Numerical Answer. Answer: 11304 to 11304

Solution

Related Formula

W = Δ U = S \cdot Δ A

For a soap bubble with two surfaces:

Δ A = 2 \times 4π (r₂² - r₁²)

Core Logic

Given: $S = 0.04 N/m$ $S = 0.04 \text{ N/m}$ Initial diameter $= 7 cm r₁ = 3.5 cm = 3.5 \times 10⁻² m$ Final diameter $= 14 cm r₂ = 7 cm = 7 \times 10⁻² m$

W = S \times (8π r₂² - 8π r₁²)

Step 1: Compute the Work Done

W = 0.04 \times 8π ((7²) - (3.5²)) \times 10⁻⁴

W = 0.04 \times 8 \times (22)/(7) \times (49 - 12.25) \times 10⁻⁴

W = 0.04 \times 2 \times (22)/(7) \times (147) \times 10⁻⁴ \times 4

Wait, the solution uses an alternate factorizing logic: $8π(49 - 12.25) = 8π(36.75) = 294π$ . Let's follow the PDF strictly:

W = 0.04 \times 2 \times (22)/(7) \times 147 \times 10⁻⁴

W = 3696 \times 10⁻⁶ J = 3696

Step 2: Solve for x

3696 = 15000 - x

x = 15000 - 3696

x = 11304

Pattern Recognition

A soap bubble has two free surfaces; always multiply area changes by 2 ( $W = 8π S Δ r²$ $W = 8\pi S \Delta r^2$). Failing to account for both surfaces yields exactly half the answer, acting as a common trap.

Chapter Mix

Class 11 Physics: Mechanical Properties of Fluids

Q4 jee_main_2025_02_april_evening Surface Tension and Surface Energy

Two water drops each of radius

r

coalesce to form a bigger drop. If

T

is the surface tension, the surface energy released in this process is:

A. $4π r² T [2 - 2(2)/(3)]$
B. $4π r² T [2 - 2(1)/(3)]$
C. $4π r² T [1 + \sqrt(2)]$ $4\pi r^2 T \left[1 + \sqrt{2}\right]$
D. $4π r² T [\sqrt(2) - 1]$ $4\pi r^2 T \left[\sqrt{2} - 1\right]$

Solution

Related Formula

Surface Energy:

U = T \cdot A = T \cdot (4π R²)

Conservation of Volume during coalescence of drops:

2 \times ((4)/(3)π r³) = (4)/(3)π R³

Core Logic

When two drops of radius $r$ $r$ coalesce into a single larger drop of radius $R$ $R$, volume is conserved:

R³ = 2r³ R = 21/3 r

Initial surface area of the two separate drops:

Aᵢ = 2 \times 4π r² = 8π r²

Final surface area of the combined single drop:

Af = 4π R² = 4π (21/3 r)² = 4π r² 22/3

Surface energy released:

Δ E = Uᵢ - Uf = T(Aᵢ - Af)

Δ E = T ( 8π r² - 4π r² 22/3 ) = 4π r² T [ 2 - 22/3 ]

Pattern Recognition

Sees: Coalescence of $N$ $N$ identical drops. Trap: Forgetting to conserve volume first, or confusing initial and final surface areas. Shortcut: Energy released when $N$ $N$ drops coalesce into one big drop is:

Δ E = 4π r² T [ N - N2/3 ]

Here, substituting $N = 2$ $N = 2$ directly gives $4π r² T [ 2 - 22/3 ]$ $4\pi r^2 T \left[ 2 - 2^{2/3} \right]$.

Chapter Mix

Class 11 Physics: Mechanical Properties of Fluids

Solution & Explanation

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Mechanical Properties of Fluids Previous-Year Questions — Page 2

Solution

Related Formula

Core Logic

Step 1: Calculation in CGS Units

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Terminal Velocity Ratio

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute the Work Done

Step 2: Solve for x

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

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