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When two soap bubbles of radius $r_{1}$ and $r_{2}$ ( $r_{2}$ > $r_{1}$ ) coalesce, the radius of curvature of common surface is

$r_{2} - r_{1}$

$\frac{r_{2} - r_{1}}{r_{2} r_{1}}$

$\frac{r_{2} r_{1}}{r_{2} - r_{1}}$

$r_{2} + r_{1}$

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Explanation

$\Rightarrow \frac{4 T}{r} = \frac{4 T}{r_{2}} - \frac{4 T}{r_{1}} \Rightarrow r = \frac{r_{1} r_{2}}{r_{1} - r_{2}}$

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The excess pressure due to surface tension in a spherical liquid drop of radius r is directly proportional to

$r^{2}$

$r^{- 1}$

$r^{- 2}$

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Explanation

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A long cylindrical glass vessel has a small hole of radius 'r' at its bottom. The depth to which the vessel can be lowered vertically in the deep water bath (surface tension T) without any water entering inside is

$4 T / rρg$

$3 T / rρg$

$2 T / rρg$

$T / rρg$

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Explanation

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The excess of pressure inside a soap bubble than that of the outer pressure is

$\frac{2 T}{r}$

$\frac{4 T}{r}$

$\frac{T}{2 r}$

$\frac{T}{r}$

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Explanation

(b) The excess of pressure inside a soap bubble = $\frac{4 T}{r}$

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The pressure of air in a soap bubble of 0.7cm diameter is 8 mm of water above the pressure outside. The surface tension of the soap solution is

100 dyne/cm

68.66 dyne/cm

137 dyne/cm

150 dyne/cm

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Explanation

(b) $∆ P = \frac{4 T}{r} = hdg \Rightarrow T = \frac{rhdg}{4} = \frac{0.35 \times 0.8 \times 1 \times 10^{3}}{4} = 70 dyne / cm \equiv 68.66 dyne / cm$

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Pressure inside two soap bubbles are 1.01 and 1.02 atmospheres. Ratio between their volumes is

102 : 101

$(102)^{3} : (101)^{3}$

8 : 1

2 : 1

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Explanation

(c) Outside pressure = 1 atm
Pressure inside first bubble = 1.01 atm
Pressure inside second bubble = 1.02 atm
Excess pressure $∆ P_{1}$ =1.01-1 = 0.01 atm
Excess pressure atm $∆ P_{2}$ = 1.02-1 = 0.02 atm

$∆ P \propto \frac{1}{r} \Rightarrow r \propto \frac{1}{∆ P} \Rightarrow \frac{r_{1}}{r_{2}} = \frac{∆ P_{2}}{∆ P_{1}} = \frac{0.02}{0.01} = \frac{2}{1}$

Since $V = \frac{4}{3} {πr}^{3} ∴ \frac{V_{1}}{V_{2}} = {(\frac{r_{1}}{r_{2}})}^{3} = {(\frac{2}{1})}^{3} = \frac{8}{1}$

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The radii of two soap bubbles are $r_{1}$ and $r_{2}$ . In isothermal conditions, two meet together in vaccum. Then the radius of the resultant bubble is given by

$R = (r_{1} + r_{2}) / 2$

$R = r_{1} (r_{1} r_{2} + r_{2})$

$R^{2} = r_{1}^{2} + r_{2}^{2}$

$R = r_{1} + r_{2}$

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Explanation

$PV = P_{1} V_{1} + P_{2} V_{2} \Rightarrow \frac{4 T}{R} \times \frac{4}{3} {πR}^{3} = \frac{4 T}{r_{1}} \times \frac{4}{3} {πr}_{1}^{3} + \frac{4 T}{r_{2}} \times \frac{4}{3} {πr}_{2}^{3} \Rightarrow R^{2} = {r_{1}}^{2} + {r_{2}}^{2}$

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When a large bubble rises from the bottom of a lake to the surface, its radius doubles. If atmospheric pressure is equal to that of column of water height H, then the depth of lake is

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Explanation

(c)

$P_{1} V_{1} = P_{2} V_{2} \Rightarrow (H + h) ρg \times \frac{4}{3} {πr}^{3} = H \times \frac{4}{3} π (2 r)^{3} \Rightarrow H + h = 8 H ∴ h = 7 H$

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Excess pressure of one soap bubble is four times more than the other. Then the ratio of volume of first bubble to another one is

1 : 64

1 : 4

64 : 1

1 : 2

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Explanation

(a) $∆ P = \frac{4 T}{r} \Rightarrow \frac{∆ P_{1}}{∆ P_{2}} = 4 ∴ \frac{r_{2}}{r_{1}} = 4 and \frac{V_{1}}{V_{2}} = {(\frac{r_{1}}{r_{2}})}^{3} = \frac{1}{64}$

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There are two liquid drops of different radii. The excess pressure inside over the outside is

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