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

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]

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(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]

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(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]

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(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]

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(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]

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(1150)

[IMAGE 87]

$Pressure inside the bubble$

$P = P_{0} + h ρ g + \frac{2 T}{r}$

$P - P_{0} = h ρ g + \frac{2 T}{r}$

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

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

Topic Question Set

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]

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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]

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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]

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