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

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

Q32 jee_main_2024_27_jan_morning Viscosity and Surface Tension

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:

A. $Statement I is correct but Statement II is incorrect$
B. $Statement I is incorrect but Statement II is correct$
C. $Both Statement I and Statement II are incorrect$
D. $Both Statement I and Statement II are correct$

Solution

Core Logic

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.

Pattern Recognition

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.

Chapter Mix

Class 11 Physics: Mechanical Properties of Fluids

Q33 jee_main_2024_29_jan_morning Surface Tension and Capillarity

Given below are two statements: Statement I: If a capillary tube is immersed first in cold water and then in hot water, the height of capillary rise will be smaller in hot water. Statement II: If a capillary tube is immersed first in cold water and then in hot water, the height of capillary rise will be smaller in cold water. In the light of the above statements, choose the most appropriate from the options given below:

A. Both Statement I and Statement II are true
B. Both Statement I and Statement II are false
C. Statement I is true but Statement II is false
D. Statement I is false but Statement II is true

Solution

Related Formula

The height of capillary rise ( $h$ $h$) is given by:

h = (2T θ)/(ρ g r)

where, $T$ $T$ = surface tension of the liquid $θ$ $\theta$ = angle of contact $ρ$ $\rho$ = density of the liquid $r$ $r$ = radius of the capillary tube

Core Logic

As the temperature of water increases, its intermolecular cohesive forces decrease. This leads to a decrease in surface tension ( $T$ $T$).

Since $h \propto T$ $h \propto T$ (assuming $ρ$ $\rho$ and $θ$ $\theta$ remain relatively constant), height of capillary rise decreases with increase in temperature.

Step 1: Evaluate Statement I and II

Because $Thot lt Tcold$ $T_{\text{hot}} \lt T_{\text{cold}}$, it follows that $hhot lt hcold$ $h_{\text{hot}} \lt h_{\text{cold}}$.

Statement I states: the height of capillary rise will be smaller in hot water. This is True.
Statement II states: the height of capillary rise will be smaller in cold water. This is False.

Pattern Recognition

Remember the key physical dependence: Temperature Up $\implies$ Surface Tension Down $\implies$ Capillary Rise Down. This basic trend of cohesive/adhesive properties versus thermal energy frequently appears in competitive conceptual physical chemistry/fluid physics questions.

Chapter Mix

Class 11 Physics: Mechanical Properties of Fluids

Q60 jee_main_2024_29_jan_morning Bernoulli's Principle

In a test experiment on a model aeroplane in wind tunnel, the flow speeds on the upper and lower surfaces of the wings are

70~ms⁻¹

and

65~ms⁻¹

respectively. If the wing area is

2~m²

the lift of the wing is ________ N. (Given density of air

= 1.2~kg m⁻³

)

Numerical Answer. Answer: 810 to 810

Solution

Related Formula

From Bernoulli's Principle (ignoring small height differences across the wing thickness):

P₁ + (1)/(2) ρ v₁² = P₂ + (1)/(2) ρ v₂²

Δ P = P₂ - P₁ = (1)/(2) ρ (v₁² - v₂²)

Net aerodynamic lift force ( $F$ $F$) acting on wing area $A$ $A$ is:

F = Δ P \cdot A = (1)/(2) ρ (v₁² - v₂²) A

Core Logic

Given values:

Flow speed on upper surface ( $v₁$ $v_1$) = $70 ~ms⁻¹$ $70 \mathrm{~ms}^{-1}$
Flow speed on lower surface ( $v₂$ $v_2$) = $65 ~ms⁻¹$ $65 \mathrm{~ms}^{-1}$
Wing area ( $A$ $A$) = $2 ~m²$ $2 \mathrm{~m}^2$
Density of air ( $ρ$ $\rho$) = $1.2 ~kg\cdot m⁻³$ $1.2 \mathrm{~kg\cdot m}^{-3}$

Step 1: Compute Lift Force

Substituting these metrics directly into the dynamic lift equation:

F = (1)/(2) \times 1.2 \times (70² - 65²) \times 2

Notice that the factors $(1)/(2)$ $\frac{1}{2}$ and $2$ $2$ cancel out perfectly:

F = 1.2 \times (4900 - 4225)

F = 1.2 \times 675 = 810 ~N

Therefore, the net lift force on the wing is $810 ~N$ $810 \mathrm{~N}$.

Pattern Recognition

Use the algebraic difference of squares shortcut $a² - b² = (a-b)(a+b)$ $a^2 - b^2 = (a-b)(a+b)$ to solve velocity square differences quickly: $70² - 65² = (70-65)(70+65) = 5 \times 135 = 675$ , bypassing squaring heavy multi-digit calculations.

Chapter Mix

Class 11 Physics: Mechanical Properties of Fluids

Q52 jee_main_2024_30_january_evening Surface Energy of Drops

A big drop is formed by coalescing

1000

small identical drops of water. If

E₁

be the total surface energy of

1000

small drops of water and

E₂

be the surface energy of single big drop of water, the

E₁:E₂

is

x:1

where

x =

Numerical Answer. Answer: 10 to 10

Solution

Related Formula

Vinitial = Vfinal

$E = S \times A$ $E = S \times A$

Core Logic

When small drops coalesce to form a large drop, the total volume is conserved.

1000 \times (4)/(3) π r³ = (4)/(3) π R³

R³ = 1000 r³ R = 10r

Step 1: Calculate Surface Energies

The total surface energy of $1000$ $1000$ small drops ( $E₁$ $\mathrm{E}_1$) is:

E₁ = 1000 \times 4π r² \times S

The surface energy of the single big drop ( $E₂$ $\mathrm{E}_2$) is:

E₂ = 4π R² \times S = 4π (10r)² \times S = 100 \times 4π r² \times S

Step 2: Find the Ratio

E₁E₂ = (1000)/(100) = (10)/(1)

Thus, the ratio is $10:1$ $10:1$, which means $x = 10$ $x = 10$.

Pattern Recognition

When $N$ $N$ droplets merge to form one big drop, the radius scales as $R = N1/3 r$ $R = N^{1/3} r$. The ratio of total initial surface energy to final surface energy is $N1/3$ $N^{1/3}$.

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 6

Solution

Related Formula

Core Logic

Step 1: Calculate the Change in Surface Area

Step 2: Calculate Work Done

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Evaluate Statement I and II

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute Lift Force

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Surface Energies

Step 2: Find the Ratio

Pattern Recognition

Chapter Mix

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