JEE Advanced Surface Tension, Surface Energy & Capillarity PYQ

Q 1 :

A glass capillary tube is of the shape of a truncated cone with an apex angle $α$ so that its two ends have cross sections of different radii. When dipped in water vertically, water rises in it to a height $h$ , where the radius of its cross section is $b$ . If the surface tension of water is $S$ , its density is $ρ$ , and its contact angle with glass is $θ$ , the value of $h$ will be ( $g$ is the acceleration due to gravity) [2014]

[IMAGE 400]

$\frac{2 S}{b ρ g} \cos (θ - α)$
$\frac{2 S}{b ρ g} \cos (θ + α)$
$\frac{2 S}{b ρ g} \cos (θ - \frac{α}{2})$
$\frac{2 S}{b ρ g} \cos (θ + \frac{α}{2})$

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(4)

Q 2 :

A spherical soap bubble inside an air chamber at pressure $P_{0} = 10^{5} Pa$ has a certain radius so that the excess pressure inside the bubble is $Δ P = 144 Pa$ . Now, the chamber pressure is reduced to $\frac{8 P_{0}}{27}$ so that the bubble radius and its excess pressure change. In this process, all the temperatures remain unchanged. Assume air to be an ideal gas and the excess pressure $Δ P$ in both the cases to be much smaller than the chamber pressure. The new excess pressure $Δ P$ in $P a$ is __________. [2024]

(96)

$Since temperature remains unchanged i.e., T = constant$

$So P_{1} V_{1} = P_{2} V_{2}$

$V_{1} = \frac{4}{3} π R_{1}^{3} and V_{2} = \frac{4}{3} π R_{2}^{3}$

$Excess pressure, P_{1} = P_{0} + Δ P_{1} and Δ P_{1} = \frac{4 T}{R_{1}}$

$P_{2} = \frac{8 P_{0}}{27} + Δ P_{2} and Δ P_{2} = \frac{4 T}{R_{2}}$

$∴$ $(P_{0} + Δ P_{1}) \times \frac{4}{3} π R_{1}^{3} = (\frac{8 P_{0}}{27} + Δ P_{2}) \times \frac{4}{3} π R_{2}^{3}$

$or, (P_{0} + Δ P_{1}) {(\frac{4 T}{Δ P_{1}})}^{3} = (\frac{8 P_{0}}{27} + Δ P_{2}) {(\frac{4 T}{Δ P_{2}})}^{3}$ $[∵ Δ P_{1} ≪ P_{0}]$

$\frac{P_{0}}{{(Δ P_{1})}^{3}} \approx \frac{8 P_{0}}{27} \times \frac{1}{{(Δ P_{2})}^{3}}$

$or, Δ P_{2} = \frac{2}{3} Δ P_{1} = \frac{2}{3} \times (144 Pa)$

$∴$ $Δ P_{2} = 96 Pa$

Q 3 :

A drop of liquid of radius $R = 10^{- 2} m$ having surface tension $S = \frac{0.1}{4 π} {Nm}^{- 1}$ divides itself into $K$ identical drops. In this process, the total change in the surface energy is $Δ U = 10^{- 3} J .$ If $K = 10^{α}$ , then the value of $α$ is _______. [2017]

(6)

$\frac{4}{3} π R^{3} = k \times \frac{4}{3} π r^{3}$ $∴$ $R = K^{1 / 3} r$

$Δ U = S [k \times 4 π r^{2} - 4 π R^{2}]$

$∴$ $Δ U = 4 π S [k \times \frac{R^{2}}{k^{2 / 3}} - R^{2}] = 4 π S R^{2} [k^{1 / 3} - 1]$

$∴$ $Δ U = 4 π S R^{2} [10^{α / 3} - 1]$ $[∵ K = 10^{α}]$

$∴$ $10^{- 3} = 4 π \times \frac{0.1}{4 π} \times {(10^{- 2})}^{2} [10^{α / 3} - 1]$

$∴$ $10^{2} = 10^{α / 3} - 1 Neglecting 1$

$10^{2} = 10^{α / 3} \Rightarrow \frac{α}{3} = 2$ $∴$ $α = 6$

Q 4 :

Two soap bubbles $A$ and $B$ are kept in a closed chamber where the air is maintained at pressure $8 {N/m}^{2}$ . The radii of bubbles $A$ and $B$ are 2 cm and 4 cm, respectively. Surface tension of the soap-water used to make bubbles is $0.04 N/m$ . Find the ratio $\frac{n_{B}}{n_{A}},$ where $n_{A}$ and $n_{B}$ are the number of moles of air in bubbles $A$ and $B$ , respectively. [Neglect the effect of gravity.] [2009]

(6)

[IMAGE 401]

$Pressure,$

$P_{A} = P_{air} + \frac{4 T}{R_{A}}$

$(∵ Excess pressure due to surface tension (T) in soap bubble = \frac{4 T}{r})$

$P_{A} = 8 + \frac{4 T}{R_{A}} = \frac{4 \times 0.04}{0.02} + 8 \Rightarrow P_{A} = 16 {N/m}^{2}$

$Similarly, P_{B} = 8 + \frac{4 T}{R_{B}} = 8 + \frac{4 \times 0.04}{0.04} = 12 {N/m}^{2}$

$According to ideal gas equation, P_{V} = n R T$

$P_{A} V_{A} = n_{A} R T_{B} \Rightarrow 16 \times \frac{4}{3} π {(0.02)}^{3} = n_{A} R T_{A} ...(i)$

$P_{B} V_{B} = n_{B} R T_{B} \Rightarrow 12 \times \frac{4}{3} π {(0.04)}^{3} = n_{B} R T_{B} ...(ii)$

$Dividing eq. (ii) by (i)$

$\frac{12 \times \frac{4}{3} π {(0.04)}^{3}}{16 \times \frac{4}{3} π {(0.02)}^{3}} = \frac{n_{B}}{n_{A}} [∵ T_{A} = T_{B}]$

$∴$ $\frac{n_{B}}{n_{A}} = 6$

Q 5 :

[IMAGE 402]

When water is filled carefully in a glass, one can fill it to a height $h$ above the rim of the glass due to the surface tension of water. To calculate $h$ just before water starts flowing, model the shape of the water above the rim as a disc of thickness $h$ having semicircular edges, as shown schematically in the figure. When the pressure of water at the bottom of this disc exceeds what can be withstood due to the surface tension, the water surface breaks near the rim and water starts flowing from there. If the density of water, $10^{3} {kg m}^{- 3}, 0.07 {Nm}^{- 1}$ and $10 {ms}^{- 2},$ respectively, the value of $h$ (in mm) is _______. [2020]

(3.74)

[IMAGE 403]

$According to question, the water above the rim is a$ $disc of thickness h having semicircular edges.$

$r = \frac{h}{2}$

$Pressure at the bottom of disc = pressure due to surface tension$

$ρ g h = T (\frac{1}{R_{1}} + \frac{1}{R_{2}})$

$R_{1} ≫ R_{2}$ $∴$ $\frac{1}{R_{1}} ≪ \frac{1}{R_{2}} and R_{2} = \frac{h}{2}$

$∴$ $ρ g h = T [\frac{1}{R_{1}} + \frac{1}{R_{2}}] = T [0 + \frac{1}{h / 2}] = \frac{2 T}{h}$

$\Rightarrow h^{2} = \frac{2 T}{ρ g} \Rightarrow h = \sqrt{\frac{2 T}{ρ g}} = \sqrt{\frac{20 \times 0.07}{10^{3} \times 10}} = \sqrt{\frac{14 \times 100}{10^{4} \times 100}}$

$∴$ $h = \sqrt{14} mm = 3.741$

Q 6 :

A cylindrical capillary tube of radius 0.2 mm is made by joining two capillaries $T_{1}$ and $T_{2}$ of different materials having water contact angles of $0 °$ and $60 °$ , respectively. The capillary tube is dipped vertically in water in two different configurations, case I and II as shown in figure. Which of the following option(s) is (are) correct?

[Surface tension of water $= 0.075 N/m$ , density of water $= 1000 {kg/m}^{3}, g = 10 {m/s}^{2}$ ] [2019]

[IMAGE 404]

The correction in the height of water column raised in the tube, due to weight of water contained in the meniscus, will be different for both cases.
For case II, if the capillary joint is 5 cm above the water surface, the height of water column raised in the tube will be 3.75 cm. (Neglect the weight of water in the meniscus).
For case I, if the joint is kept at 8 cm above the water surface, the height of water column in the tube will be 7.5 cm. (Neglect the weight of the water in the meniscus.)
For case I, if the capillary joint is 5 cm above the water surface, the height of water column raised in the tube will be more than 8.75 cm. (Neglect the weight of water in the meniscus.)

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Select one or more options

(1, 2, 3)

$For case I$

$h_{1} = \frac{2 T \cos θ_{1}}{r ρ g} = \frac{2 \times 0.75 \times \cos 0 °}{2 \times 10^{- 4} \times 1000 \times 10} = 7.5 cm$

$For case II$

$h_{2} = \frac{2 T \cos θ_{2}}{r ρ g} = \frac{2 \times 0.75 \times \cos 60 °}{2 \times 10^{- 4} \times 1000 \times 10} = 3.75 cm$

The correction in the height of water column raised in the tube, due to weight of water contained in the meniscus will be different for both cases.

In case II, if the capillary joint is 5 cm above the water surface then water in capillary will not reach the interface.

Water will reach only till 3.75 cm.

Q 7 :

A uniform capillary tube of inner radius $r$ is dipped vertically into a beaker filled with water. The water rises to a height $h$ in the capillary tube above the water surface in the beaker. The surface tension of water is $σ$ . The angle of contact between water and the wall of the capillary tube is $θ$ . Ignore the mass of water in the meniscus. Which of the following statements is (are) true? [2018]

For a given material of the capillary tube, $h$ decreases with increase in $r$
For a given material of the capillary tube, $h$ is independent of $σ$
If this experiment is performed in a lift going up with a constant acceleration, then $h$ decreases
$h$ is proportional to contact angle $θ$

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Select one or more options

(1, 3)

$As we know h = \frac{2 σ \cos θ}{r ρ g_{eff}}$

$As' r' increases, h decreases h \propto \frac{1}{r}$

[all other parameters remaining constant]

$Also h \propto σ$

Further if lift is going up with an acceleration $' a'$

$then g_{eff} = g + a . As g_{eff} increases, h decreases.$

$Also h \propto \cos θ not h \propto θ$

Q 8 :

When liquid medicine of density $ρ$ is to put in the eye, it is done with the help of a dropper. As the bulb on the top of the dropper is pressed, a drop forms at the opening of the dropper. We wish to estimate the size of the drop. We first assume that the drop formed at the opening is spherical because that requires a minimum increase in its surface energy. To determine the size, we calculate the net vertical force due to the surface tension T when the radius of the drop is R. When this force becomes smaller than the weight of the drop, the drop gets detached from the dropper.

Q. If the radius of the opening of the dropper is $r$ , the vertical force due to the surface tension on the drop of radius $R$ (assuming $r ≪ R$ ) is [2010]

$2 π r T$
$2 π R T$
$\frac{2 π r^{2} T}{R}$
$\frac{2 π R^{2} T}{r}$

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(3)

[IMAGE 405]

Vertical force due to surface tension

$F_{V} = (T \cos θ) \times 2 π r$

$= T (\frac{r}{R}) \times 2 π r = \frac{2 π r^{2} T}{R}$

$[∵ \cos θ = \frac{r}{R}]$

Q 9 :

When liquid medicine of density $ρ$ is to put in the eye, it is done with the help of a dropper. As the bulb on the top of the dropper is pressed, a drop forms at the opening of the dropper. We wish to estimate the size of the drop. We first assume that the drop formed at the opening is spherical because that requires a minimum increase in its surface energy. To determine the size, we calculate the net vertical force due to the surface tension T when the radius of the drop is R. When this force becomes smaller than the weight of the drop, the drop gets detached from the dropper.

Q. If $r = 5 \times 10^{- 4} m, ρ = 10^{3} {kgm}^{- 3}, g = 10 {ms}^{- 2}, T = 0.11 {Nm}^{- 1},$ the radius of the drop when it detaches from the dropper is approximately [2010]

$1.4 \times 10^{- 3} m$
$3.3 \times 10^{- 3} m$
$2.0 \times 10^{- 3} m$
$4.1 \times 10^{- 3} m$

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(1)

When the drop is about to detach from the dropper

$Weight = vertical force due to surface tension (m g = F_{v})$

$∴$ $\frac{4}{3} π R^{3} ρ g = \frac{2 π r^{2} T}{R}$

$\Rightarrow R^{4} = (\frac{3}{2} \frac{r^{2} T}{ρ g}) = \frac{3}{2} \times \frac{{(5 \times 10^{- 4})}^{2} \times 0.11}{1000 \times 10} = 4.12 \times 10^{- 12}$

$or, R = 1.42 \times 10^{- 3} m$

Q 10 :

When liquid medicine of density $ρ$ is to put in the eye, it is done with the help of a dropper. As the bulb on the top of the dropper is pressed, a drop forms at the opening of the dropper. We wish to estimate the size of the drop. We first assume that the drop formed at the opening is spherical because that requires a minimum increase in its surface energy. To determine the size, we calculate the net vertical force due to the surface tension T when the radius of the drop is R. When this force becomes smaller than the weight of the drop, the drop gets detached from the dropper.

Q. After the drop detaches, its surface energy is [2010]

$1.4 \times 10^{- 6} J$
$2.7 \times 10^{- 6} J$
$5.4 \times 10^{- 6} J$
$8.1 \times 10^{- 6} J$

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(2)

$Surface energy = T \times 4 π R^{2}$

$= 0.11 \times 4 \times \frac{22}{7} \times {(1.42 \times 10^{- 3})}^{2} = 2.7 \times 10^{- 6} J$

Topic Question Set

(4)

Q 3 :

A drop of liquid of radius $R = 10^{- 2} m$ having surface tension $S = \frac{0.1}{4 π} {Nm}^{- 1}$ divides itself into $K$ identical drops. In this process, the total change in the surface energy is $Δ U = 10^{- 3} J .$ If $K = 10^{α}$ , then the value of $α$ is _______. [2017]

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Topic Question Set

(4)

Q 3 : A drop of liquid of radius R=10-2 m having surface tension S=0.14π Nm-1 divides itself into K identical drops. In this process, the total change in the surface energy is ΔU=10-3 J. If K=10α, then the value of α is _______. [2017]

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Q 3 :

A drop of liquid of radius $R = 10^{- 2} m$ having surface tension $S = \frac{0.1}{4 π} {Nm}^{- 1}$ divides itself into $K$ identical drops. In this process, the total change in the surface energy is $Δ U = 10^{- 3} J .$ If $K = 10^{α}$ , then the value of $α$ is _______. [2017]