The amount of work done to break a big water drop of radius 'R' into 27 small drops of equal radius is 10 J. The work…

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

The amount of work done to break a big water drop of radius 'R' into 27 small drops of equal radius is 10 J. The work done required to break the same big drop into 64 small drops of equal radius will be :-

Solution & Explanation

Related Formula

Core Logic

Step 1: Set up Proportions

Pattern Recognition

Chapter Mix

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

More Mechanical Properties of Fluids Questions — jee_main_2025_24_jan_morning

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