Fluid Mechanics | Surface Tension And Viscosity | JEE Main previous year questions with Solutions

Q 1 :

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: [2024]

Statement I is correct but statement II is incorrect
Statement I is incorrect but Statement II is correct
Both Statement I and Statement II are incorrect
Both Statement I and Statement II are correct

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(2)

Gases have less viscosity.

Due to insoluble impurities like detergent, surface tension decreases.

Q 2 :

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 [2024]

$4 {πR}^{2} T$
$8 {πR}^{2} T$
$3 {πR}^{2} T$
$1 / 8 {πR}^{2} T$

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(2)

Volume of liquid of drops should be constant,

$\frac{4}{3} {πR}^{3} = 27 \times \frac{4}{3} \times {πr}^{3}$

$R^{3} = 27 r^{3} \Rightarrow R = 3 r$

$r = \frac{R}{3} \Rightarrow r^{2} = \frac{R^{2}}{9}$

Work done = $T \cdot ∆ A$

$= 27 T (4 {πr}^{2}) - T 4 {πR}^{2}$

$= 27 T 4 π \frac{R^{2}}{9} - 4 {πR}^{2} T = 8 {πR}^{2} T$

Q 3 :

A big drop is formed by coalescing 1000 small droplets of water. The surface energy will become [2024]

100 times
10 times
$\frac{1}{100}$ th
$\frac{1}{10}$ th

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(4)

Lets say radius of small droplets is $r$ and that of big drop is R,

By conservation of volume:

$\frac{4}{3} {πR}^{3} = 1000 \frac{4}{3} {πr}^{3} \Rightarrow R = 10 r$

$∴$ Surface energy of 1000 droplets

$U_{i} = 1000 (4 {πr}^{2} S)$

$U_{f} = 4 π R^{2} = 100 (4 {πr}^{2} S)$

$U_{f} = \frac{1}{10} U_{i}$

Q 4 :

A soap bubble is blown to a diameter of 7 cm. 36960 erg of work is done in blowing it further. If surface tension of soap solution is 40 dyne/cm then the new radius is _______ cm. Take $(π = \frac{22}{7})$ [2024]

(7) $Δ W = 2 T \cdot 4 π r_{2}^{2} - 2 T \cdot 4 π r_{1}^{2}$

$r_{2}^{2} - r_{1}^{2} = \frac{Δ W}{8 T π} = \frac{36960 \times 7}{8 \times 40 \times 22} = 36.75$

$r_{2}^{2} = 36.75 + \frac{49}{4} = 49 \Rightarrow r_{2} = 7 cm$

Q 5 :

A big drop is formed by coalescing 1000 small identical drops of water. If $E_{1}$ be the total surface energy of 1000 small drops of water and $E_{2}$ be the surface energy of single big drop of water, the $E_{1} : E_{0}$ is $x : 1$ where $x =$ __________ . [2024]

(10) Let radius of small drops = r

Let radius of big drops = R

Here volume will remain conserve (volume conservation)

$∴ (1000) V_{s} = (1) V_{b} \to volume of big drop$

$Volume of small drop \Rightarrow 10^{3} (\frac{4}{3} π r^{3}) = \frac{4}{3} π R^{3}$

$R = 10 r$

$We know, surface tension = \frac{Surface energy}{Area}$

$∴ Surface energy = (Surface tension) (Area)$

$\Rightarrow ε_{1} = T \times 4 π r^{2} \times 10^{3}$

$ε_{2} = T \times 4 π R^{2}$

$∴ \frac{ε_{1}}{ε_{2}} = \frac{(10^{3}) (r^{2})}{R^{2}} = \frac{(10^{3}) r^{2}}{{(10 r)}^{2}} = \frac{10}{1}$

$\frac{ε_{1}}{ε_{2}} = \frac{10}{1} = \frac{X}{1},$

$∴ x = 10$

Q 6 :

A big drop is formed by coalescing 1000 small droplets of water. The ratio of surface energy of 1000 droplets to that of the energy of the big drop is $\frac{10}{x}$ . The value of $x$ is ____. [2024]

(1)

$1000 \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3} \Rightarrow R = 10 r$

Surface energy of 1000 drops $= 1000 \times T \times 4 π r^{2}$

Surface energy of bigger drop $= T \times 4 π R^{2}$

$\frac{S.E. of 1000 drops}{S.E. of Big drop} = \frac{1000 (4 π r^{2}) T}{4 π R^{2} T} = \frac{1000 \times r^{2}}{{(10 r)}^{2}}$

$\Rightarrow \frac{S.E. of 1000 drops}{S.E. of Big drop} = 10 = \frac{10}{x}$

$∴$ $x = 1$

Q 7 :

An air bubble of radius 0.1 cm lies at a depth of 20 cm below the free surface of a liquid of density 1000 $kg / m^{3}$ . If the pressure inside the bubble is 2100 $N / m^{2}$ greater than the atmospheric pressure, then the surface tension of the liquid in SI unit is (use g = 10 $m / s^{2}$ ) [2025]

0.02
0.1
0.25
0.05

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(4)

$P_{1} = P_{0} + p g h = P_{0} + 1000 \times 10 \times \frac{20}{100}$

$\Rightarrow P_{1} = P_{0} + 2000$

So, $P_{2} - P_{1} = \frac{2 T}{R} = (\frac{2 T}{1 \times 10^{- 3}})$

$\Rightarrow P_{2} = P_{0} + 2100$ (given)

So, $P_{0} + 2100 - P_{0} - 2000 = 2 T \times 10^{3}$

$\Rightarrow 100 = 2 T \times 10^{3}$

$\Rightarrow T = (\frac{1}{20}) = 0.05 N / m$

Q 8 :

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 [2025]

15 J
10 J
20 J
5 J

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(1)

The amount of work done to break a big water drop into small drops,

$W = △ U = T △ A$

One drop to n drops

$\frac{4}{3} π R^{3} = n \frac{4}{3} π r^{3} \Rightarrow r = \frac{R}{n^{\frac{1}{3}}}$

So, $W = T (n 4 π r^{2} - 4 π R^{2}) = T 4 π R^{2} (n^{1 / 3} - 1)$

For one drop to 27 drops

$W = T 4 π R^{2} (27^{1 / 3} - 1)$ ...(i)

For one drop to 64 drops

$W' = T 4 π R^{2} (64^{1 / 3} - 1)$ ...(ii)

(ii)/(i)

$\frac{W'}{W} = \frac{4 - 1}{3 - 1} = \frac{3}{2} \Rightarrow W' = \frac{3}{2} W = 15 J$

Q 9 :

Consider following statements:

A. Surface tension arises due to extra energy of the molecules at the interior as compared to the molecules at the surface, of a liquid.

B. As the temperature of liquid rises, the coefficient of viscosity increases.

C. As the temperature of gas increases, the coefficient of viscosity increases.

D. The onset of turbulence is determined by Reynold's number.

E. In a steady flow two stream lines never intersect.

Choose the correct answer from the options given below: [2025]

A, D, E only
C, D, E only
B, C, D only
A, B, C only

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(2)

Surface tension arises due to extra energy of the molecules at the surface as compared to the interior of a liquid. The coefficient of viscosity for a liquid decreases with rise in temperature where as it increases for gases with increase in temperature. The flow is turbulent for a Reynold's number greater than 2000. Stream lines never intersect in a steady flow.

Q 10 :

Two water drops each of radius 'r' coalesce to from a bigger drop. If 'T' is the surface tension, the surface energy released in this process is: [2025]

$4 π r^{2} T [2 - 2^{\frac{2}{3}}]$
$4 π r^{2} T [2 - 2^{\frac{1}{3}}]$
$4 π r^{2} T [1 + \sqrt{2}]$
$4 π r^{2} T [\sqrt{2} - 1]$

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(1)

$2 \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3} \Rightarrow R = 2^{1 / 3} r$

$E_{i} = 2 \times 4 π r^{2} T$

$E_{f} = 4 π R^{2} T = 4 π r^{2} 2^{2 / 3} T$

Heat lost $△ Q = E_{i} - E_{f} = 4 π r^{2} T [2 - 2^{2 / 3}]$

Q 11 :

Two soap bubbles of radius 2 cm and 4 cm, respectively, are in contact with each other. The radius of curvature of the common surface, in cm, is __________. [2025]

(4)

The excess pressure inside a soap bubble is given by the formula,

$△ P = \frac{4 T}{R}$ , where

T is the surface tension of the soap solution, and R is the radius of the bubble.

For the first bubble ( $R_{1}$ = 2 cm) : $△ P_{1} = \frac{4 T}{R_{1}}$

For the first bubble ( $R_{2}$ = 4 cm) : $△ P_{2} = \frac{4 T}{R_{2}}$

When the two bubbles join, the difference in pressure between the two bubbles will equal the pressure difference across the interface (R):

$△ P_{1} - △ P_{2} = △ P$

$\Rightarrow \frac{4 T}{R_{1}} - \frac{4 T}{R_{2}} = \frac{4 T}{R}$

$\Rightarrow R = \frac{R_{1} \cdot R_{2}}{R_{2} - R_{1}} = \frac{2 \cdot 4}{4 - 2} = 4 cm$

Q 12 :

An air bubble of radius 1.0 mm is observed at a depth of 20 cm below the free surface of a liquid having surface tension 0.095 $J / m^{2}$ and density $10^{3} kg / m^{3}$ . The difference between pressure inside the bubble and atmospheric pressure ________ $N / m^{2}$ . (Take g = 10 $m / s^{2}$ ) [2025]

(2190)

$△ P = P_{in} - P_{0}$

$= ρ g h + \frac{2 T}{R} = \frac{1000 \times 10 \times 20}{100} + \frac{2 \times 0.095}{10^{- 3}}$

$= 2000 + 190 = 2190 Pa$

Q 13 :

Surface tension of a soap bubble is $2.0 \times 10^{- 2} {Nm}^{- 1}$ . Work done to increase the radius of the soap bubble from 3.5 cm to 7 cm will be [Take $π = \frac{22}{7}$ ] [2023]

$0.72 \times 10^{- 4} J$
$5.76 \times 10^{- 4} J$
$18.48 \times 10^{- 4} J$
$9.24 \times 10^{- 4} J$

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(3)

$Surface area of soap bubble = 2 \times 4 π R^{2}$

$Work done = change in surface energy \times T_{s}$

$= T_{s} \times 8 π \times (R_{2}^{2} - R_{1}^{2})$

$= 2 \times 10^{- 2} \times 8 \times \frac{22}{7} \times 49 \times \frac{3}{4} \times 10^{- 4}$

$= 18.48 \times 10^{- 4} J$

Q 14 :

If 1000 droplets of water of surface tension 0.07 N/m, having same radius 1 mm each, combine to form a single drop, In the process the released surface energy is (Take $π = \frac{22}{7}$ ) [2023]

$7.92 \times 10^{- 6} J$
$7.92 \times 10^{- 4} J$
$9.68 \times 10^{- 4} J$
$8.8 \times 10^{- 5} J$

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(2)

$1000 \times \frac{4 π}{3} {(1)}^{3} = \frac{4 π}{3} R^{3}$

$R = 10 mm$

$T \times 1000 \times 4 π {(10^{- 3})}^{2} - T \times 4 π {(10 \times 10^{- 3})}^{2} = Δ E$

$Δ E = 4 \times π \times 7 \times 10^{- 2} [1000 - 100] \times 10^{- 6}$

$Δ E = 7.92 \times 10^{- 4} J$

Q 15 :

A mercury drop of radius $10^{- 3} m$ is broken into 125 equal size droplets. Surface tension of mercury is $0.45 {Nm}^{- 1}$ . The gain in surface energy is [2023]

$2.26 \times 10^{- 5} J$
$28 \times 10^{- 5} J$
$17.5 \times 10^{- 5} J$
$5 \times 10^{- 5} J$

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(1)

$Initial surface energy = 0.45 \times 4 π {(10^{- 3})}^{2}$

$\frac{4}{3} π {(10^{- 3})}^{2} = 125 \times \frac{4 π}{3} R_{new}^{2}$

$∴$ $10^{- 3} = 5 R_{new}$

$∴$ $R_{new} = \frac{10^{- 3}}{5} m$

$So, final surface energy = 0.45 \times 125 \times 4 π {(\frac{10^{- 3}}{5})}^{2}$

$Increase in energy = 0.45 \times 4 π \times {(10^{- 3})}^{2} [\frac{125}{25} - 1]$

$= 4 \times 0.45 \times 4 π \times 10^{- 6} = 2.26 \times 10^{- 5} J$

Q 16 :

A spherical drop of liquid splits into 1000 identical spherical drops. If $u_{i}$ is the surface energy of the original drop and $u_{f}$ is the total surface energy of the resulting drops, the (ignoring evaporation) $\frac{u_{f}}{u_{i}} = (\frac{10}{x}) .$ Then value of $x$ is ___________ . [2023]

(1)

$Surface tension = T$

$R : Radius of bigger drop$

$r : Radius of smaller drop$

$Volume will remain same$

$\frac{4}{3} π R^{3} = 1000 \times \frac{4}{3} π r^{3}$ $\Rightarrow R = 10 r$

$U_{i} = T \cdot 4 π R^{2}$

$U_{f} = T \cdot 4 π r^{2} \times 1000$

$\frac{U_{f}}{U_{i}} = \frac{1000 r^{2}}{R^{2}} = \frac{10}{1}$

$So, x = 1$

Q 17 :

The surface tension of soap solution is $3.5 \times 10^{- 2} {Nm}^{- 1}$ . The amount of work done required to increase the radius of a soap bubble from 10 cm to 20 cm is ______ $\times 10^{- 4} J .$ [2023]

(264)

$W = T (Δ A)$

$W = T (8 π (r_{2}^{2} - r_{1}^{2}))$ $= 264 \times 10^{- 4} J$

Q 18 :

There is an air bubble of radius 1.0 mm in a liquid of surface tension 0.075 ${Nm}^{- 1}$ and density 1000 ${kg m}^{- 3}$ at a depth of 10 cm below the free surface. The amount by which the pressure inside the bubble is greater than the atmospheric pressure is ________ Pa $(g = 10 {ms}^{- 2})$ [2023]

(1150)

$Pressure inside the bubble$

$P = P_{0} + h ρ g + \frac{2 T}{r}$

$P - P_{0} = h ρ g + \frac{2 T}{r}$

$= 0.1 \times 1000 \times 10 + \frac{2 \times 0.075}{10^{- 3}}$

$= 1000 + (0.1) (1000)$ $= 1150 Pa$

Q 19 :

Given below are two statements:

Statement I: Pressure of a fluid is exerted only on a solid surface in contact as the fluid-pressure does not exist everywhere in a still fluid.

Statement II: Excess potential energy of the molecules on the surface of a liquid, when compared to interior, results in surface tension.

In the light of the above statements, choose the correct answer from the options given below. [2026]

Both Statement I and Statement II are true
Statement I is true but Statement II is false
Both Statement I and Statement II are false
Statement I is false but Statement II is true

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(4)

According to Pascal’s law, pressure at any point in a liquid at rest is the same in all directions.

It exists at every point in the liquid, not just at the boundaries. So statement (1) is false.

For an interior molecule, the net cohesive forces are zero. Hence statement (2) is correct.

Q 20 :

A soap bubble of surface tension 0.04 N/m is blown to a diameter of 7 cm. If $(15000 - x) μ J$ of work is done in blowing it further to make its diameter 14 cm, then the value of $x$ is _______. $(π = 22 / 7)$ [2026]

(11304)

$W = Δ u$

$= S \times (8 π r_{2}^{2} - 8 π r_{1}^{2})$

$= 0.04 \times 2 \times \frac{22}{7} (147) \times 10^{- 4}$

$W = 3696 \times 10^{- 6} J$

$3696 = 15000 - x$

$x = 11304 μ J$

Q 21 :

When a part of a straight capillary tube is placed vertically in a liquid, the liquid raises upto certain height h. If the inner radius of the capillary tube, density of the liquid and surface tension of the liquid decrease by 1% each, then the height of the liquid in the tube will change by ______ %. [2026]

+3
-1
+1
-3

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(3)

$h = \frac{2 T \cos θ}{ρ g r}$

$\frac{Δ h}{h} % = \frac{Δ T}{T} % - \frac{Δ ρ}{ρ} % - \frac{Δ r}{r} %$

$\frac{Δ h}{h} % = 1 + 1 + 1$

$\frac{Δ h}{h} = + 1 %$

Topic Question Set

Q 2 :

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 [2024]

Q 3 :

A big drop is formed by coalescing 1000 small droplets of water. The surface energy will become [2024]

Q 4 :

A soap bubble is blown to a diameter of 7 cm. 36960 erg of work is done in blowing it further. If surface tension of soap solution is 40 dyne/cm then the new radius is _______ cm. Take $(π = \frac{22}{7})$ [2024]

Q 5 :

A big drop is formed by coalescing 1000 small identical drops of water. If $E_{1}$ be the total surface energy of 1000 small drops of water and $E_{2}$ be the surface energy of single big drop of water, the $E_{1} : E_{0}$ is $x : 1$ where $x =$ __________ . [2024]

Q 6 :

A big drop is formed by coalescing 1000 small droplets of water. The ratio of surface energy of 1000 droplets to that of the energy of the big drop is $\frac{10}{x}$ . The value of $x$ is ____. [2024]

Q 8 :

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 [2025]

Q 10 :

Two water drops each of radius 'r' coalesce to from a bigger drop. If 'T' is the surface tension, the surface energy released in this process is: [2025]

Q 11 :

Two soap bubbles of radius 2 cm and 4 cm, respectively, are in contact with each other. The radius of curvature of the common surface, in cm, is __________. [2025]

Q 13 :

Surface tension of a soap bubble is $2.0 \times 10^{- 2} {Nm}^{- 1}$ . Work done to increase the radius of the soap bubble from 3.5 cm to 7 cm will be [Take $π = \frac{22}{7}$ ] [2023]

Q 14 :

If 1000 droplets of water of surface tension 0.07 N/m, having same radius 1 mm each, combine to form a single drop, In the process the released surface energy is (Take $π = \frac{22}{7}$ ) [2023]

Q 15 :

A mercury drop of radius $10^{- 3} m$ is broken into 125 equal size droplets. Surface tension of mercury is $0.45 {Nm}^{- 1}$ . The gain in surface energy is [2023]

Q 17 :

The surface tension of soap solution is $3.5 \times 10^{- 2} {Nm}^{- 1}$ . The amount of work done required to increase the radius of a soap bubble from 10 cm to 20 cm is ______ $\times 10^{- 4} J .$ [2023]

Q 20 :

A soap bubble of surface tension 0.04 N/m is blown to a diameter of 7 cm. If $(15000 - x) μ J$ of work is done in blowing it further to make its diameter 14 cm, then the value of $x$ is _______. $(π = 22 / 7)$ [2026]

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Topic Question Set

Q 2 : 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 [2024]

Q 3 : A big drop is formed by coalescing 1000 small droplets of water. The surface energy will become [2024]

Q 4 : A soap bubble is blown to a diameter of 7 cm. 36960 erg of work is done in blowing it further. If surface tension of soap solution is 40 dyne/cm then the new radius is _______ cm. Take (π=227) [2024]

Q 5 : 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:E0 is x:1 where x= __________ . [2024]

Q 6 : A big drop is formed by coalescing 1000 small droplets of water. The ratio of surface energy of 1000 droplets to that of the energy of the big drop is 10x. The value of x is ____. [2024]

Q 7 : An air bubble of radius 0.1 cm lies at a depth of 20 cm below the free surface of a liquid of density 1000 kg/m3. If the pressure inside the bubble is 2100 N/m2 greater than the atmospheric pressure, then the surface tension of the liquid in SI unit is (use g = 10 m/s2) [2025]

Q 8 : 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 [2025]

Q 10 : Two water drops each of radius 'r' coalesce to from a bigger drop. If 'T' is the surface tension, the surface energy released in this process is: [2025]

Q 11 : Two soap bubbles of radius 2 cm and 4 cm, respectively, are in contact with each other. The radius of curvature of the common surface, in cm, is __________. [2025]

Q 12 : An air bubble of radius 1.0 mm is observed at a depth of 20 cm below the free surface of a liquid having surface tension 0.095 J/m2 and density 103 kg/m3. The difference between pressure inside the bubble and atmospheric pressure ________ N/m2. (Take g = 10 m/s2) [2025]

Q 13 : Surface tension of a soap bubble is 2.0×10-2 Nm-1. Work done to increase the radius of the soap bubble from 3.5 cm to 7 cm will be [Take π=227] [2023]

Q 14 : If 1000 droplets of water of surface tension 0.07 N/m, having same radius 1 mm each, combine to form a single drop, In the process the released surface energy is (Take π=227) [2023]

Q 15 : A mercury drop of radius 10-3 m is broken into 125 equal size droplets. Surface tension of mercury is 0.45 Nm-1. The gain in surface energy is [2023]

Q 16 : A spherical drop of liquid splits into 1000 identical spherical drops. If ui is the surface energy of the original drop and uf is the total surface energy of the resulting drops, the (ignoring evaporation) ufui=(10x). Then value of x is ___________ . [2023]

Q 17 : The surface tension of soap solution is 3.5×10-2Nm-1. The amount of work done required to increase the radius of a soap bubble from 10 cm to 20 cm is ______ ×10-4 J. [2023]

Q 18 : There is an air bubble of radius 1.0 mm in a liquid of surface tension 0.075 Nm-1and density 1000 kg m-3 at a depth of 10 cm below the free surface. The amount by which the pressure inside the bubble is greater than the atmospheric pressure is ________ Pa (g=10ms-2) [2023]

Q 20 : 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=22/7) [2026]

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Q 2 :

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 [2024]

Q 3 :

A big drop is formed by coalescing 1000 small droplets of water. The surface energy will become [2024]

Q 4 :

A soap bubble is blown to a diameter of 7 cm. 36960 erg of work is done in blowing it further. If surface tension of soap solution is 40 dyne/cm then the new radius is _______ cm. Take $(π = \frac{22}{7})$ [2024]

Q 5 :

A big drop is formed by coalescing 1000 small identical drops of water. If $E_{1}$ be the total surface energy of 1000 small drops of water and $E_{2}$ be the surface energy of single big drop of water, the $E_{1} : E_{0}$ is $x : 1$ where $x =$ __________ . [2024]

Q 6 :

A big drop is formed by coalescing 1000 small droplets of water. The ratio of surface energy of 1000 droplets to that of the energy of the big drop is $\frac{10}{x}$ . The value of $x$ is ____. [2024]

Q 8 :

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 [2025]

Q 10 :

Two water drops each of radius 'r' coalesce to from a bigger drop. If 'T' is the surface tension, the surface energy released in this process is: [2025]

Q 11 :

Two soap bubbles of radius 2 cm and 4 cm, respectively, are in contact with each other. The radius of curvature of the common surface, in cm, is __________. [2025]

Q 13 :

Surface tension of a soap bubble is $2.0 \times 10^{- 2} {Nm}^{- 1}$ . Work done to increase the radius of the soap bubble from 3.5 cm to 7 cm will be [Take $π = \frac{22}{7}$ ] [2023]

Q 14 :

If 1000 droplets of water of surface tension 0.07 N/m, having same radius 1 mm each, combine to form a single drop, In the process the released surface energy is (Take $π = \frac{22}{7}$ ) [2023]

Q 15 :

A mercury drop of radius $10^{- 3} m$ is broken into 125 equal size droplets. Surface tension of mercury is $0.45 {Nm}^{- 1}$ . The gain in surface energy is [2023]

Q 17 :

The surface tension of soap solution is $3.5 \times 10^{- 2} {Nm}^{- 1}$ . The amount of work done required to increase the radius of a soap bubble from 10 cm to 20 cm is ______ $\times 10^{- 4} J .$ [2023]

Q 20 :

A soap bubble of surface tension 0.04 N/m is blown to a diameter of 7 cm. If $(15000 - x) μ J$ of work is done in blowing it further to make its diameter 14 cm, then the value of $x$ is _______. $(π = 22 / 7)$ [2026]