QBManager

Mechanical Properties of Fluids

143058 The air pressure inside a soap bubble of radius $R$ exceeds the outside air pressure by $10 Pa$ . By how much will the pressure inside a bubble of radius $2 R$ exceed the outside air pressure?

1

20 Pa

2

40 Pa

3

2.5 Pa

4

5 Pa

Explanation:

D Given, $P_{1} = 10 pa, R_{1} = R, R_{2} = 2 R$
Excess pressure inside a soap bubble,
$P = \frac{4 T}{R}$
Where, $T$ is the surface tension of soap solution and $R$ is the radius of the bubble.
$P \propto \frac{1}{R}$
$∴ \frac{P_{1}}{P_{2}} = \frac{R_{2}}{R_{1}}$
$P_{2} = \frac{P_{1} R_{1}}{R_{2}} = \frac{R \times 10}{2 R}$
$P_{2} = 5 Pa$

J and K-CET-2012

Mechanical Properties of Fluids

143059 64 identical water droplets combine to form a large drop. Find the ratio of the total surface energy of 64 droplets to that of the large drop

1

64 : 1

2

4 : 1

3

8 : 1

4

1 : 8

Explanation:

B Identical water droplets combine to form a large drop
$\frac{4}{3} π R^{3} = 64 \times \frac{4}{3} π r^{3}$
$R = 4 r$
Ratio of the total surface energy of 64 droplets to that of the large drop
$= \frac{(S.E.)_{I}}{(S.E.)_{L}} = \frac{64 \times 4 π r^{2}}{4 π R^{2}}$
Putting the value of $R$ from equation (i), we get
$\frac{(S.E.)_{I}}{(S.E.)_{L}} = \frac{64 \times 4 π r^{2}}{4 π \times 16 r^{2}}$
$= \frac{4}{1}$
$\frac{(S.E.)_{I}}{(S.E.)_{L}} = 4 : 1$

J and K-CET-2017

Mechanical Properties of Fluids

143060 When a large bubble rises from bottom of a water lake to its surface, then its radius doubles. If the atmospheric pressure is equal to the pressure of height ' $H$ ' of a certain water column, then the depth of lake will be

1

H

2

4 H

3

7 H

4

2 H

Explanation:

C :-The radius of a large bubble is double, when it travels from bottom of the lake to its surface. The pressure of height ' $H$ ' of a certain water column is equal to atmospheric pressure.
Volume $(V) = \frac{4}{3} π r^{3}$
Where, $r$ is the radius of the bubble.
From equation of Boyle's law,
$P_{1} V_{1} = P_{2} V_{2}$
$P_{atm} \frac{4}{3} π (2 r)^{3} = (P_{atm} + ρ g h) \times \frac{4}{3} π r^{3}$
$P_{atm} \times 8 = P_{atm} + ρ g h$
$H ρ g \times 8 = H ρ g + ρ gh$
$7 H ρ g = ρ gh$
$h = \frac{7 H ρ g}{ρ g}$
$h = 7 H$

MHT-CET 2019

Mechanical Properties of Fluids

143061 A drop of some liquid of volume $0.04 {cm}^{3}$ is placed on the surface of a glass slide. Then another glass slide is placed on it in such a way that the liquid forms a thin layer of area $20 {cm}^{2}$ between the surfaces of the two slides. To separate the slides a force of $16 \times 10^{5}$ dyne has to be applied normal to the surfaces. The surface tension of the liquid is (in dyne-cm $^{- 1}$ )

1 60

2 70

3 80

4 90

Explanation:

C Given that, volume (V) $= 0.04 {cm}^{3}$ ,
Area $(A) = 20 {cm}^{2}$ , Force $(F) = 16 \times 10^{5}$ dyne

We know that
Volume $(V) = A \times l$
$l = \frac{V}{A} = \frac{0.04}{20} = 0.002 cm$
$∵ \frac{l}{2} = r$
$r = \frac{l}{2} = \frac{0.002}{2} = 0.001 cm$
Excess pressure $(P) = T [\frac{1}{R} - \frac{1}{r}]$
$(∵ R = \infty)$
$P = \frac{T}{r}$
$F / A = \frac{T}{r}$
$T = \frac{Fr}{A}$
$T = \frac{16 \times 10^{5} \times 0.001}{20}$
Surface tension, $T = 80$ dyne- ${cm}^{- 1}$

WB JEE 2014

Mechanical Properties of Fluids

143058 The air pressure inside a soap bubble of radius $R$ exceeds the outside air pressure by $10 Pa$ . By how much will the pressure inside a bubble of radius $2 R$ exceed the outside air pressure?

1

20 Pa

2

40 Pa

3

2.5 Pa

4

5 Pa

Explanation:

D Given, $P_{1} = 10 pa, R_{1} = R, R_{2} = 2 R$
Excess pressure inside a soap bubble,
$P = \frac{4 T}{R}$
Where, $T$ is the surface tension of soap solution and $R$ is the radius of the bubble.
$P \propto \frac{1}{R}$
$∴ \frac{P_{1}}{P_{2}} = \frac{R_{2}}{R_{1}}$
$P_{2} = \frac{P_{1} R_{1}}{R_{2}} = \frac{R \times 10}{2 R}$
$P_{2} = 5 Pa$

J and K-CET-2012

Mechanical Properties of Fluids

143059 64 identical water droplets combine to form a large drop. Find the ratio of the total surface energy of 64 droplets to that of the large drop

1

64 : 1

2

4 : 1

3

8 : 1

4

1 : 8

Explanation:

B Identical water droplets combine to form a large drop
$\frac{4}{3} π R^{3} = 64 \times \frac{4}{3} π r^{3}$
$R = 4 r$
Ratio of the total surface energy of 64 droplets to that of the large drop
$= \frac{(S.E.)_{I}}{(S.E.)_{L}} = \frac{64 \times 4 π r^{2}}{4 π R^{2}}$
Putting the value of $R$ from equation (i), we get
$\frac{(S.E.)_{I}}{(S.E.)_{L}} = \frac{64 \times 4 π r^{2}}{4 π \times 16 r^{2}}$
$= \frac{4}{1}$
$\frac{(S.E.)_{I}}{(S.E.)_{L}} = 4 : 1$

J and K-CET-2017

Mechanical Properties of Fluids

143060 When a large bubble rises from bottom of a water lake to its surface, then its radius doubles. If the atmospheric pressure is equal to the pressure of height ' $H$ ' of a certain water column, then the depth of lake will be

1

H

2

4 H

3

7 H

4

2 H

Explanation:

C :-The radius of a large bubble is double, when it travels from bottom of the lake to its surface. The pressure of height ' $H$ ' of a certain water column is equal to atmospheric pressure.
Volume $(V) = \frac{4}{3} π r^{3}$
Where, $r$ is the radius of the bubble.
From equation of Boyle's law,
$P_{1} V_{1} = P_{2} V_{2}$
$P_{atm} \frac{4}{3} π (2 r)^{3} = (P_{atm} + ρ g h) \times \frac{4}{3} π r^{3}$
$P_{atm} \times 8 = P_{atm} + ρ g h$
$H ρ g \times 8 = H ρ g + ρ gh$
$7 H ρ g = ρ gh$
$h = \frac{7 H ρ g}{ρ g}$
$h = 7 H$

MHT-CET 2019

Mechanical Properties of Fluids

143061 A drop of some liquid of volume $0.04 {cm}^{3}$ is placed on the surface of a glass slide. Then another glass slide is placed on it in such a way that the liquid forms a thin layer of area $20 {cm}^{2}$ between the surfaces of the two slides. To separate the slides a force of $16 \times 10^{5}$ dyne has to be applied normal to the surfaces. The surface tension of the liquid is (in dyne-cm $^{- 1}$ )

1 60

2 70

3 80

4 90

Explanation:

C Given that, volume (V) $= 0.04 {cm}^{3}$ ,
Area $(A) = 20 {cm}^{2}$ , Force $(F) = 16 \times 10^{5}$ dyne

We know that
Volume $(V) = A \times l$
$l = \frac{V}{A} = \frac{0.04}{20} = 0.002 cm$
$∵ \frac{l}{2} = r$
$r = \frac{l}{2} = \frac{0.002}{2} = 0.001 cm$
Excess pressure $(P) = T [\frac{1}{R} - \frac{1}{r}]$
$(∵ R = \infty)$
$P = \frac{T}{r}$
$F / A = \frac{T}{r}$
$T = \frac{Fr}{A}$
$T = \frac{16 \times 10^{5} \times 0.001}{20}$
Surface tension, $T = 80$ dyne- ${cm}^{- 1}$

WB JEE 2014

Mechanical Properties of Fluids

143058 The air pressure inside a soap bubble of radius $R$ exceeds the outside air pressure by $10 Pa$ . By how much will the pressure inside a bubble of radius $2 R$ exceed the outside air pressure?

1

20 Pa

2

40 Pa

3

2.5 Pa

4

5 Pa

Explanation:

D Given, $P_{1} = 10 pa, R_{1} = R, R_{2} = 2 R$
Excess pressure inside a soap bubble,
$P = \frac{4 T}{R}$
Where, $T$ is the surface tension of soap solution and $R$ is the radius of the bubble.
$P \propto \frac{1}{R}$
$∴ \frac{P_{1}}{P_{2}} = \frac{R_{2}}{R_{1}}$
$P_{2} = \frac{P_{1} R_{1}}{R_{2}} = \frac{R \times 10}{2 R}$
$P_{2} = 5 Pa$

J and K-CET-2012

Mechanical Properties of Fluids

143059 64 identical water droplets combine to form a large drop. Find the ratio of the total surface energy of 64 droplets to that of the large drop

1

64 : 1

2

4 : 1

3

8 : 1

4

1 : 8

Explanation:

B Identical water droplets combine to form a large drop
$\frac{4}{3} π R^{3} = 64 \times \frac{4}{3} π r^{3}$
$R = 4 r$
Ratio of the total surface energy of 64 droplets to that of the large drop
$= \frac{(S.E.)_{I}}{(S.E.)_{L}} = \frac{64 \times 4 π r^{2}}{4 π R^{2}}$
Putting the value of $R$ from equation (i), we get
$\frac{(S.E.)_{I}}{(S.E.)_{L}} = \frac{64 \times 4 π r^{2}}{4 π \times 16 r^{2}}$
$= \frac{4}{1}$
$\frac{(S.E.)_{I}}{(S.E.)_{L}} = 4 : 1$

J and K-CET-2017

Mechanical Properties of Fluids

143060 When a large bubble rises from bottom of a water lake to its surface, then its radius doubles. If the atmospheric pressure is equal to the pressure of height ' $H$ ' of a certain water column, then the depth of lake will be

1

H

2

4 H

3

7 H

4

2 H

Explanation:

C :-The radius of a large bubble is double, when it travels from bottom of the lake to its surface. The pressure of height ' $H$ ' of a certain water column is equal to atmospheric pressure.
Volume $(V) = \frac{4}{3} π r^{3}$
Where, $r$ is the radius of the bubble.
From equation of Boyle's law,
$P_{1} V_{1} = P_{2} V_{2}$
$P_{atm} \frac{4}{3} π (2 r)^{3} = (P_{atm} + ρ g h) \times \frac{4}{3} π r^{3}$
$P_{atm} \times 8 = P_{atm} + ρ g h$
$H ρ g \times 8 = H ρ g + ρ gh$
$7 H ρ g = ρ gh$
$h = \frac{7 H ρ g}{ρ g}$
$h = 7 H$

MHT-CET 2019

Mechanical Properties of Fluids

143061 A drop of some liquid of volume $0.04 {cm}^{3}$ is placed on the surface of a glass slide. Then another glass slide is placed on it in such a way that the liquid forms a thin layer of area $20 {cm}^{2}$ between the surfaces of the two slides. To separate the slides a force of $16 \times 10^{5}$ dyne has to be applied normal to the surfaces. The surface tension of the liquid is (in dyne-cm $^{- 1}$ )

1 60

2 70

3 80

4 90

Explanation:

C Given that, volume (V) $= 0.04 {cm}^{3}$ ,
Area $(A) = 20 {cm}^{2}$ , Force $(F) = 16 \times 10^{5}$ dyne

We know that
Volume $(V) = A \times l$
$l = \frac{V}{A} = \frac{0.04}{20} = 0.002 cm$
$∵ \frac{l}{2} = r$
$r = \frac{l}{2} = \frac{0.002}{2} = 0.001 cm$
Excess pressure $(P) = T [\frac{1}{R} - \frac{1}{r}]$
$(∵ R = \infty)$
$P = \frac{T}{r}$
$F / A = \frac{T}{r}$
$T = \frac{Fr}{A}$
$T = \frac{16 \times 10^{5} \times 0.001}{20}$
Surface tension, $T = 80$ dyne- ${cm}^{- 1}$

WB JEE 2014

Mechanical Properties of Fluids

143058 The air pressure inside a soap bubble of radius $R$ exceeds the outside air pressure by $10 Pa$ . By how much will the pressure inside a bubble of radius $2 R$ exceed the outside air pressure?

1

20 Pa

2

40 Pa

3

2.5 Pa

4

5 Pa

Explanation:

D Given, $P_{1} = 10 pa, R_{1} = R, R_{2} = 2 R$
Excess pressure inside a soap bubble,
$P = \frac{4 T}{R}$
Where, $T$ is the surface tension of soap solution and $R$ is the radius of the bubble.
$P \propto \frac{1}{R}$
$∴ \frac{P_{1}}{P_{2}} = \frac{R_{2}}{R_{1}}$
$P_{2} = \frac{P_{1} R_{1}}{R_{2}} = \frac{R \times 10}{2 R}$
$P_{2} = 5 Pa$

J and K-CET-2012

Mechanical Properties of Fluids

143059 64 identical water droplets combine to form a large drop. Find the ratio of the total surface energy of 64 droplets to that of the large drop

1

64 : 1

2

4 : 1

3

8 : 1

4

1 : 8

Explanation:

B Identical water droplets combine to form a large drop
$\frac{4}{3} π R^{3} = 64 \times \frac{4}{3} π r^{3}$
$R = 4 r$
Ratio of the total surface energy of 64 droplets to that of the large drop
$= \frac{(S.E.)_{I}}{(S.E.)_{L}} = \frac{64 \times 4 π r^{2}}{4 π R^{2}}$
Putting the value of $R$ from equation (i), we get
$\frac{(S.E.)_{I}}{(S.E.)_{L}} = \frac{64 \times 4 π r^{2}}{4 π \times 16 r^{2}}$
$= \frac{4}{1}$
$\frac{(S.E.)_{I}}{(S.E.)_{L}} = 4 : 1$

J and K-CET-2017

Mechanical Properties of Fluids

143060 When a large bubble rises from bottom of a water lake to its surface, then its radius doubles. If the atmospheric pressure is equal to the pressure of height ' $H$ ' of a certain water column, then the depth of lake will be

1

H

2

4 H

3

7 H

4

2 H

Explanation:

C :-The radius of a large bubble is double, when it travels from bottom of the lake to its surface. The pressure of height ' $H$ ' of a certain water column is equal to atmospheric pressure.
Volume $(V) = \frac{4}{3} π r^{3}$
Where, $r$ is the radius of the bubble.
From equation of Boyle's law,
$P_{1} V_{1} = P_{2} V_{2}$
$P_{atm} \frac{4}{3} π (2 r)^{3} = (P_{atm} + ρ g h) \times \frac{4}{3} π r^{3}$
$P_{atm} \times 8 = P_{atm} + ρ g h$
$H ρ g \times 8 = H ρ g + ρ gh$
$7 H ρ g = ρ gh$
$h = \frac{7 H ρ g}{ρ g}$
$h = 7 H$

MHT-CET 2019

Mechanical Properties of Fluids

143061 A drop of some liquid of volume $0.04 {cm}^{3}$ is placed on the surface of a glass slide. Then another glass slide is placed on it in such a way that the liquid forms a thin layer of area $20 {cm}^{2}$ between the surfaces of the two slides. To separate the slides a force of $16 \times 10^{5}$ dyne has to be applied normal to the surfaces. The surface tension of the liquid is (in dyne-cm $^{- 1}$ )

1 60

2 70

3 80

4 90

Explanation:

C Given that, volume (V) $= 0.04 {cm}^{3}$ ,
Area $(A) = 20 {cm}^{2}$ , Force $(F) = 16 \times 10^{5}$ dyne

We know that
Volume $(V) = A \times l$
$l = \frac{V}{A} = \frac{0.04}{20} = 0.002 cm$
$∵ \frac{l}{2} = r$
$r = \frac{l}{2} = \frac{0.002}{2} = 0.001 cm$
Excess pressure $(P) = T [\frac{1}{R} - \frac{1}{r}]$
$(∵ R = \infty)$
$P = \frac{T}{r}$
$F / A = \frac{T}{r}$
$T = \frac{Fr}{A}$
$T = \frac{16 \times 10^{5} \times 0.001}{20}$
Surface tension, $T = 80$ dyne- ${cm}^{- 1}$

WB JEE 2014