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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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