QBManager

Mechanical Properties of Fluids

143027 The work done in blowing a soap bubble of surface tension $0.06 {Nm}^{- 1}$ from $2 cm$ radius to 5 cm radius is

1

0.004168 J

2

0.003168 J

3

0.003158 J

4

0.004568 J

Explanation:

B Given,
Surface Tension $(T) = 0.06 {Nm}^{- 1}$
Radius of soap bubbles
$R_{1} = 2 cm = 2 \times 10^{- 2} m$
$R_{2} = 5 cm = 5 \times 10^{- 2} m$
So, Surface energy (a) $= T \times$ Surface area
$V = T \times 4 π R^{2}$
$∵$ Soap bubble has 2 -free surface.
$∴$ surface energy for soap bubble $= 2 V$
$= 2 T \times 4 π R^{2}$
$= 8 π {TR}^{2}$
Now,
Work done in blowing a soap bubble from radius $2 cm$ to $5 cm =$ Increase in surface energy for soap bubble
$W = 8 π T (R_{2}^{2} - R_{1}^{2}) = 8 π T [{(5 \times 10^{- 2})}^{2} - {(2 \times 10^{- 2})}^{2}]$
$W = 8 π T (5^{2} - 2^{2}) \times 10^{- 4}$
$W = 8 π T (25 - 4) \times 10^{- 4}$
$W = 8 π T (21 \times 10^{- 4}) = 8 \times 3.14 \times 0.06 \times 21 \times 10^{- 4}$
$W = 31.65 \times 10^{- 4} J$
$W = 0.003165 J$

BITSAT-2014

Mechanical Properties of Fluids

143029 When two soap bubbles of radius $r_{1}$ and $r_{2} (r_{2}$ $> r_{1}$ ) coalesce, the radius of curvature of common surface is

1

r_{2} - r_{1}

2

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

3

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

4

r_{2} + r_{1}

Explanation:

C Given,
Radius of bubble $1 = r_{1}$
Radius of bubble $2 = r_{2}$
For bubbles 1,
$P_{1} - P_{atm} = \frac{4 T}{r_{1}}$
$P_{1} = \frac{4 T}{r_{1}} + P_{atm}$
For bubble 2,
$P_{2} - P_{atm} = \frac{4 T}{r_{2}}$
$P_{2} = \frac{4 T}{r_{2}} + P_{atm}$
Consider that two bubbles are merged,
$P_{1} - P_{2} = \frac{4 T}{r}$
$\frac{4 T}{r_{1}} + P_{atm} - \frac{4 T}{r_{2}} - P_{atm} = \frac{4 T}{r}$
$(\frac{1}{r_{1}} - \frac{1}{r_{2}}) = \frac{1}{r}$
$r = \frac{r_{1} r_{2}}{r_{2} - r_{1}}$

JCECE-2004

Mechanical Properties of Fluids

143030 The amount of work done in blowing a soap bubble such that its diameter increases from $d$ to $D$ is (T-surface tension of the solution)

1

4 π (D^{2} - d^{2}) T

2

8 π (D^{2} - d^{2}) T

3

π (D^{2} - d^{2}) T

4

2 π (D^{2} - d^{2}) T

Explanation:

D We know that,
$New surface area = π D^{2}$
$Old surface area = π d^{2}$
Increase in area of soap bubble $(Δ A) = 2 (π D^{2} - π d^{2})$
$= 2 π (D^{2} - d^{2})$
Work done $(W) =$ Increase in area of soap bubble $\times$ Surface tension
$W = 2 π (D^{2} - d^{2}) T$

CG PET- 2004

Mechanical Properties of Fluids

143031 Consider an air bubble of radius $2 mm$ in a liquid at a depth of $5 cm$ below the free surface. The density of the liquid is $1000 kg / m^{3}$ and the surface tension is $0.1 N / m$ . Then, find the pressure inside the air bubble is greater than the pressure at the free surface of the liquid. (Take, $g = 10 m / s^{2}$ )

1

500 Pa

2

600 Pa

3

700 Pa

4

800 Pa

Explanation:

B Given,
Radius of bubble $(r) = 2 mm = 0.002 m$
Depth of bubble below the water surface $(h) = 5 cm$ $= 0.05 m$
Density of liquid $(ρ) = 1000 kg / m^{3}$
Surface tension of liquid $(σ) = 0.1 N / m$
$g = 10 m / \sec^{2}$
Gauge pressure due to liquid as a depth of $5 cm -$
$P = ρ gh$
$P = 1000 \times 10 \times 0.05$
$P = 500 Pa$
Excess pressure inside the air bubble w.r.t. outside due to surface tension -
$P^{'} = \frac{2 σ}{r}$
$P^{'} = \frac{2 \times 0.1}{0.002}$
$P^{'} = 100 Pa$
Total excess pressure inside the bubble w.r.t. to surface liquid -
$P_{total} = P + P^{'}$
$P_{total} = 500 + 100$
$P_{total} = 600 Pa$

TS- EAMCET-07.05.2018

Mechanical Properties of Fluids

143032 Consider a water droplet of diameter $0.2 mm$ where the outside pressure is $1.5 N / {cm}^{2}$ at $25^{\circ} C$ . the pressure inside the droplet, when the surface tension at $25^{\circ} C$ is $0.08 N / m$ is

1

0.32 N / {cm}^{2}

2

1.18 N / {cm}^{2}

3

1.82 N / {cm}^{2}

4

1.66 N / {cm}^{2}

Explanation:

D Given,
Diameter of the water droplet $(d) = 0.2 mm$
Radius of the water droplet $r = 0.1 mm = 0.0001 m$
Pressure outside droplet $(P_{0}) = 1.5 N / {cm}^{2}$
$= 1.5 \times 10^{4} N / m^{2}$
Surface tension $(σ) = 0.08 N / m$ We know,
Pressure inside droplet $(P) = \frac{2 σ}{r} + P_{0}$
$P = \frac{2 \times 0.08}{0.0001} + 1.5 \times 10^{4}$
$P = 0.16 \times 10^{4} + 1.5 \times 10^{4}$
$P = 1.66 N / {cm}^{2}$

TS- EAMCET-06.05.2019

Mechanical Properties of Fluids

143027 The work done in blowing a soap bubble of surface tension $0.06 {Nm}^{- 1}$ from $2 cm$ radius to 5 cm radius is

1

0.004168 J

2

0.003168 J

3

0.003158 J

4

0.004568 J

Explanation:

B Given,
Surface Tension $(T) = 0.06 {Nm}^{- 1}$
Radius of soap bubbles
$R_{1} = 2 cm = 2 \times 10^{- 2} m$
$R_{2} = 5 cm = 5 \times 10^{- 2} m$
So, Surface energy (a) $= T \times$ Surface area
$V = T \times 4 π R^{2}$
$∵$ Soap bubble has 2 -free surface.
$∴$ surface energy for soap bubble $= 2 V$
$= 2 T \times 4 π R^{2}$
$= 8 π {TR}^{2}$
Now,
Work done in blowing a soap bubble from radius $2 cm$ to $5 cm =$ Increase in surface energy for soap bubble
$W = 8 π T (R_{2}^{2} - R_{1}^{2}) = 8 π T [{(5 \times 10^{- 2})}^{2} - {(2 \times 10^{- 2})}^{2}]$
$W = 8 π T (5^{2} - 2^{2}) \times 10^{- 4}$
$W = 8 π T (25 - 4) \times 10^{- 4}$
$W = 8 π T (21 \times 10^{- 4}) = 8 \times 3.14 \times 0.06 \times 21 \times 10^{- 4}$
$W = 31.65 \times 10^{- 4} J$
$W = 0.003165 J$

BITSAT-2014

Mechanical Properties of Fluids

143029 When two soap bubbles of radius $r_{1}$ and $r_{2} (r_{2}$ $> r_{1}$ ) coalesce, the radius of curvature of common surface is

1

r_{2} - r_{1}

2

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

3

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

4

r_{2} + r_{1}

Explanation:

C Given,
Radius of bubble $1 = r_{1}$
Radius of bubble $2 = r_{2}$
For bubbles 1,
$P_{1} - P_{atm} = \frac{4 T}{r_{1}}$
$P_{1} = \frac{4 T}{r_{1}} + P_{atm}$
For bubble 2,
$P_{2} - P_{atm} = \frac{4 T}{r_{2}}$
$P_{2} = \frac{4 T}{r_{2}} + P_{atm}$
Consider that two bubbles are merged,
$P_{1} - P_{2} = \frac{4 T}{r}$
$\frac{4 T}{r_{1}} + P_{atm} - \frac{4 T}{r_{2}} - P_{atm} = \frac{4 T}{r}$
$(\frac{1}{r_{1}} - \frac{1}{r_{2}}) = \frac{1}{r}$
$r = \frac{r_{1} r_{2}}{r_{2} - r_{1}}$

JCECE-2004

Mechanical Properties of Fluids

143030 The amount of work done in blowing a soap bubble such that its diameter increases from $d$ to $D$ is (T-surface tension of the solution)

1

4 π (D^{2} - d^{2}) T

2

8 π (D^{2} - d^{2}) T

3

π (D^{2} - d^{2}) T

4

2 π (D^{2} - d^{2}) T

Explanation:

D We know that,
$New surface area = π D^{2}$
$Old surface area = π d^{2}$
Increase in area of soap bubble $(Δ A) = 2 (π D^{2} - π d^{2})$
$= 2 π (D^{2} - d^{2})$
Work done $(W) =$ Increase in area of soap bubble $\times$ Surface tension
$W = 2 π (D^{2} - d^{2}) T$

CG PET- 2004

Mechanical Properties of Fluids

143031 Consider an air bubble of radius $2 mm$ in a liquid at a depth of $5 cm$ below the free surface. The density of the liquid is $1000 kg / m^{3}$ and the surface tension is $0.1 N / m$ . Then, find the pressure inside the air bubble is greater than the pressure at the free surface of the liquid. (Take, $g = 10 m / s^{2}$ )

1

500 Pa

2

600 Pa

3

700 Pa

4

800 Pa

Explanation:

B Given,
Radius of bubble $(r) = 2 mm = 0.002 m$
Depth of bubble below the water surface $(h) = 5 cm$ $= 0.05 m$
Density of liquid $(ρ) = 1000 kg / m^{3}$
Surface tension of liquid $(σ) = 0.1 N / m$
$g = 10 m / \sec^{2}$
Gauge pressure due to liquid as a depth of $5 cm -$
$P = ρ gh$
$P = 1000 \times 10 \times 0.05$
$P = 500 Pa$
Excess pressure inside the air bubble w.r.t. outside due to surface tension -
$P^{'} = \frac{2 σ}{r}$
$P^{'} = \frac{2 \times 0.1}{0.002}$
$P^{'} = 100 Pa$
Total excess pressure inside the bubble w.r.t. to surface liquid -
$P_{total} = P + P^{'}$
$P_{total} = 500 + 100$
$P_{total} = 600 Pa$

TS- EAMCET-07.05.2018

Mechanical Properties of Fluids

143032 Consider a water droplet of diameter $0.2 mm$ where the outside pressure is $1.5 N / {cm}^{2}$ at $25^{\circ} C$ . the pressure inside the droplet, when the surface tension at $25^{\circ} C$ is $0.08 N / m$ is

1

0.32 N / {cm}^{2}

2

1.18 N / {cm}^{2}

3

1.82 N / {cm}^{2}

4

1.66 N / {cm}^{2}

Explanation:

D Given,
Diameter of the water droplet $(d) = 0.2 mm$
Radius of the water droplet $r = 0.1 mm = 0.0001 m$
Pressure outside droplet $(P_{0}) = 1.5 N / {cm}^{2}$
$= 1.5 \times 10^{4} N / m^{2}$
Surface tension $(σ) = 0.08 N / m$ We know,
Pressure inside droplet $(P) = \frac{2 σ}{r} + P_{0}$
$P = \frac{2 \times 0.08}{0.0001} + 1.5 \times 10^{4}$
$P = 0.16 \times 10^{4} + 1.5 \times 10^{4}$
$P = 1.66 N / {cm}^{2}$

TS- EAMCET-06.05.2019

Mechanical Properties of Fluids

143027 The work done in blowing a soap bubble of surface tension $0.06 {Nm}^{- 1}$ from $2 cm$ radius to 5 cm radius is

1

0.004168 J

2

0.003168 J

3

0.003158 J

4

0.004568 J

Explanation:

B Given,
Surface Tension $(T) = 0.06 {Nm}^{- 1}$
Radius of soap bubbles
$R_{1} = 2 cm = 2 \times 10^{- 2} m$
$R_{2} = 5 cm = 5 \times 10^{- 2} m$
So, Surface energy (a) $= T \times$ Surface area
$V = T \times 4 π R^{2}$
$∵$ Soap bubble has 2 -free surface.
$∴$ surface energy for soap bubble $= 2 V$
$= 2 T \times 4 π R^{2}$
$= 8 π {TR}^{2}$
Now,
Work done in blowing a soap bubble from radius $2 cm$ to $5 cm =$ Increase in surface energy for soap bubble
$W = 8 π T (R_{2}^{2} - R_{1}^{2}) = 8 π T [{(5 \times 10^{- 2})}^{2} - {(2 \times 10^{- 2})}^{2}]$
$W = 8 π T (5^{2} - 2^{2}) \times 10^{- 4}$
$W = 8 π T (25 - 4) \times 10^{- 4}$
$W = 8 π T (21 \times 10^{- 4}) = 8 \times 3.14 \times 0.06 \times 21 \times 10^{- 4}$
$W = 31.65 \times 10^{- 4} J$
$W = 0.003165 J$

BITSAT-2014

Mechanical Properties of Fluids

143029 When two soap bubbles of radius $r_{1}$ and $r_{2} (r_{2}$ $> r_{1}$ ) coalesce, the radius of curvature of common surface is

1

r_{2} - r_{1}

2

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

3

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

4

r_{2} + r_{1}

Explanation:

C Given,
Radius of bubble $1 = r_{1}$
Radius of bubble $2 = r_{2}$
For bubbles 1,
$P_{1} - P_{atm} = \frac{4 T}{r_{1}}$
$P_{1} = \frac{4 T}{r_{1}} + P_{atm}$
For bubble 2,
$P_{2} - P_{atm} = \frac{4 T}{r_{2}}$
$P_{2} = \frac{4 T}{r_{2}} + P_{atm}$
Consider that two bubbles are merged,
$P_{1} - P_{2} = \frac{4 T}{r}$
$\frac{4 T}{r_{1}} + P_{atm} - \frac{4 T}{r_{2}} - P_{atm} = \frac{4 T}{r}$
$(\frac{1}{r_{1}} - \frac{1}{r_{2}}) = \frac{1}{r}$
$r = \frac{r_{1} r_{2}}{r_{2} - r_{1}}$

JCECE-2004

Mechanical Properties of Fluids

143030 The amount of work done in blowing a soap bubble such that its diameter increases from $d$ to $D$ is (T-surface tension of the solution)

1

4 π (D^{2} - d^{2}) T

2

8 π (D^{2} - d^{2}) T

3

π (D^{2} - d^{2}) T

4

2 π (D^{2} - d^{2}) T

Explanation:

D We know that,
$New surface area = π D^{2}$
$Old surface area = π d^{2}$
Increase in area of soap bubble $(Δ A) = 2 (π D^{2} - π d^{2})$
$= 2 π (D^{2} - d^{2})$
Work done $(W) =$ Increase in area of soap bubble $\times$ Surface tension
$W = 2 π (D^{2} - d^{2}) T$

CG PET- 2004

Mechanical Properties of Fluids

143031 Consider an air bubble of radius $2 mm$ in a liquid at a depth of $5 cm$ below the free surface. The density of the liquid is $1000 kg / m^{3}$ and the surface tension is $0.1 N / m$ . Then, find the pressure inside the air bubble is greater than the pressure at the free surface of the liquid. (Take, $g = 10 m / s^{2}$ )

1

500 Pa

2

600 Pa

3

700 Pa

4

800 Pa

Explanation:

B Given,
Radius of bubble $(r) = 2 mm = 0.002 m$
Depth of bubble below the water surface $(h) = 5 cm$ $= 0.05 m$
Density of liquid $(ρ) = 1000 kg / m^{3}$
Surface tension of liquid $(σ) = 0.1 N / m$
$g = 10 m / \sec^{2}$
Gauge pressure due to liquid as a depth of $5 cm -$
$P = ρ gh$
$P = 1000 \times 10 \times 0.05$
$P = 500 Pa$
Excess pressure inside the air bubble w.r.t. outside due to surface tension -
$P^{'} = \frac{2 σ}{r}$
$P^{'} = \frac{2 \times 0.1}{0.002}$
$P^{'} = 100 Pa$
Total excess pressure inside the bubble w.r.t. to surface liquid -
$P_{total} = P + P^{'}$
$P_{total} = 500 + 100$
$P_{total} = 600 Pa$

TS- EAMCET-07.05.2018

Mechanical Properties of Fluids

143032 Consider a water droplet of diameter $0.2 mm$ where the outside pressure is $1.5 N / {cm}^{2}$ at $25^{\circ} C$ . the pressure inside the droplet, when the surface tension at $25^{\circ} C$ is $0.08 N / m$ is

1

0.32 N / {cm}^{2}

2

1.18 N / {cm}^{2}

3

1.82 N / {cm}^{2}

4

1.66 N / {cm}^{2}

Explanation:

D Given,
Diameter of the water droplet $(d) = 0.2 mm$
Radius of the water droplet $r = 0.1 mm = 0.0001 m$
Pressure outside droplet $(P_{0}) = 1.5 N / {cm}^{2}$
$= 1.5 \times 10^{4} N / m^{2}$
Surface tension $(σ) = 0.08 N / m$ We know,
Pressure inside droplet $(P) = \frac{2 σ}{r} + P_{0}$
$P = \frac{2 \times 0.08}{0.0001} + 1.5 \times 10^{4}$
$P = 0.16 \times 10^{4} + 1.5 \times 10^{4}$
$P = 1.66 N / {cm}^{2}$

TS- EAMCET-06.05.2019

Mechanical Properties of Fluids

143027 The work done in blowing a soap bubble of surface tension $0.06 {Nm}^{- 1}$ from $2 cm$ radius to 5 cm radius is

1

0.004168 J

2

0.003168 J

3

0.003158 J

4

0.004568 J

Explanation:

B Given,
Surface Tension $(T) = 0.06 {Nm}^{- 1}$
Radius of soap bubbles
$R_{1} = 2 cm = 2 \times 10^{- 2} m$
$R_{2} = 5 cm = 5 \times 10^{- 2} m$
So, Surface energy (a) $= T \times$ Surface area
$V = T \times 4 π R^{2}$
$∵$ Soap bubble has 2 -free surface.
$∴$ surface energy for soap bubble $= 2 V$
$= 2 T \times 4 π R^{2}$
$= 8 π {TR}^{2}$
Now,
Work done in blowing a soap bubble from radius $2 cm$ to $5 cm =$ Increase in surface energy for soap bubble
$W = 8 π T (R_{2}^{2} - R_{1}^{2}) = 8 π T [{(5 \times 10^{- 2})}^{2} - {(2 \times 10^{- 2})}^{2}]$
$W = 8 π T (5^{2} - 2^{2}) \times 10^{- 4}$
$W = 8 π T (25 - 4) \times 10^{- 4}$
$W = 8 π T (21 \times 10^{- 4}) = 8 \times 3.14 \times 0.06 \times 21 \times 10^{- 4}$
$W = 31.65 \times 10^{- 4} J$
$W = 0.003165 J$

BITSAT-2014

Mechanical Properties of Fluids

143029 When two soap bubbles of radius $r_{1}$ and $r_{2} (r_{2}$ $> r_{1}$ ) coalesce, the radius of curvature of common surface is

1

r_{2} - r_{1}

2

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

3

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

4

r_{2} + r_{1}

Explanation:

C Given,
Radius of bubble $1 = r_{1}$
Radius of bubble $2 = r_{2}$
For bubbles 1,
$P_{1} - P_{atm} = \frac{4 T}{r_{1}}$
$P_{1} = \frac{4 T}{r_{1}} + P_{atm}$
For bubble 2,
$P_{2} - P_{atm} = \frac{4 T}{r_{2}}$
$P_{2} = \frac{4 T}{r_{2}} + P_{atm}$
Consider that two bubbles are merged,
$P_{1} - P_{2} = \frac{4 T}{r}$
$\frac{4 T}{r_{1}} + P_{atm} - \frac{4 T}{r_{2}} - P_{atm} = \frac{4 T}{r}$
$(\frac{1}{r_{1}} - \frac{1}{r_{2}}) = \frac{1}{r}$
$r = \frac{r_{1} r_{2}}{r_{2} - r_{1}}$

JCECE-2004

Mechanical Properties of Fluids

143030 The amount of work done in blowing a soap bubble such that its diameter increases from $d$ to $D$ is (T-surface tension of the solution)

1

4 π (D^{2} - d^{2}) T

2

8 π (D^{2} - d^{2}) T

3

π (D^{2} - d^{2}) T

4

2 π (D^{2} - d^{2}) T

Explanation:

D We know that,
$New surface area = π D^{2}$
$Old surface area = π d^{2}$
Increase in area of soap bubble $(Δ A) = 2 (π D^{2} - π d^{2})$
$= 2 π (D^{2} - d^{2})$
Work done $(W) =$ Increase in area of soap bubble $\times$ Surface tension
$W = 2 π (D^{2} - d^{2}) T$

CG PET- 2004

Mechanical Properties of Fluids

143031 Consider an air bubble of radius $2 mm$ in a liquid at a depth of $5 cm$ below the free surface. The density of the liquid is $1000 kg / m^{3}$ and the surface tension is $0.1 N / m$ . Then, find the pressure inside the air bubble is greater than the pressure at the free surface of the liquid. (Take, $g = 10 m / s^{2}$ )

1

500 Pa

2

600 Pa

3

700 Pa

4

800 Pa

Explanation:

B Given,
Radius of bubble $(r) = 2 mm = 0.002 m$
Depth of bubble below the water surface $(h) = 5 cm$ $= 0.05 m$
Density of liquid $(ρ) = 1000 kg / m^{3}$
Surface tension of liquid $(σ) = 0.1 N / m$
$g = 10 m / \sec^{2}$
Gauge pressure due to liquid as a depth of $5 cm -$
$P = ρ gh$
$P = 1000 \times 10 \times 0.05$
$P = 500 Pa$
Excess pressure inside the air bubble w.r.t. outside due to surface tension -
$P^{'} = \frac{2 σ}{r}$
$P^{'} = \frac{2 \times 0.1}{0.002}$
$P^{'} = 100 Pa$
Total excess pressure inside the bubble w.r.t. to surface liquid -
$P_{total} = P + P^{'}$
$P_{total} = 500 + 100$
$P_{total} = 600 Pa$

TS- EAMCET-07.05.2018

Mechanical Properties of Fluids

143032 Consider a water droplet of diameter $0.2 mm$ where the outside pressure is $1.5 N / {cm}^{2}$ at $25^{\circ} C$ . the pressure inside the droplet, when the surface tension at $25^{\circ} C$ is $0.08 N / m$ is

1

0.32 N / {cm}^{2}

2

1.18 N / {cm}^{2}

3

1.82 N / {cm}^{2}

4

1.66 N / {cm}^{2}

Explanation:

D Given,
Diameter of the water droplet $(d) = 0.2 mm$
Radius of the water droplet $r = 0.1 mm = 0.0001 m$
Pressure outside droplet $(P_{0}) = 1.5 N / {cm}^{2}$
$= 1.5 \times 10^{4} N / m^{2}$
Surface tension $(σ) = 0.08 N / m$ We know,
Pressure inside droplet $(P) = \frac{2 σ}{r} + P_{0}$
$P = \frac{2 \times 0.08}{0.0001} + 1.5 \times 10^{4}$
$P = 0.16 \times 10^{4} + 1.5 \times 10^{4}$
$P = 1.66 N / {cm}^{2}$

TS- EAMCET-06.05.2019

Mechanical Properties of Fluids

143027 The work done in blowing a soap bubble of surface tension $0.06 {Nm}^{- 1}$ from $2 cm$ radius to 5 cm radius is

1

0.004168 J

2

0.003168 J

3

0.003158 J

4

0.004568 J

Explanation:

B Given,
Surface Tension $(T) = 0.06 {Nm}^{- 1}$
Radius of soap bubbles
$R_{1} = 2 cm = 2 \times 10^{- 2} m$
$R_{2} = 5 cm = 5 \times 10^{- 2} m$
So, Surface energy (a) $= T \times$ Surface area
$V = T \times 4 π R^{2}$
$∵$ Soap bubble has 2 -free surface.
$∴$ surface energy for soap bubble $= 2 V$
$= 2 T \times 4 π R^{2}$
$= 8 π {TR}^{2}$
Now,
Work done in blowing a soap bubble from radius $2 cm$ to $5 cm =$ Increase in surface energy for soap bubble
$W = 8 π T (R_{2}^{2} - R_{1}^{2}) = 8 π T [{(5 \times 10^{- 2})}^{2} - {(2 \times 10^{- 2})}^{2}]$
$W = 8 π T (5^{2} - 2^{2}) \times 10^{- 4}$
$W = 8 π T (25 - 4) \times 10^{- 4}$
$W = 8 π T (21 \times 10^{- 4}) = 8 \times 3.14 \times 0.06 \times 21 \times 10^{- 4}$
$W = 31.65 \times 10^{- 4} J$
$W = 0.003165 J$

BITSAT-2014

Mechanical Properties of Fluids

143029 When two soap bubbles of radius $r_{1}$ and $r_{2} (r_{2}$ $> r_{1}$ ) coalesce, the radius of curvature of common surface is

1

r_{2} - r_{1}

2

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

3

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

4

r_{2} + r_{1}

Explanation:

C Given,
Radius of bubble $1 = r_{1}$
Radius of bubble $2 = r_{2}$
For bubbles 1,
$P_{1} - P_{atm} = \frac{4 T}{r_{1}}$
$P_{1} = \frac{4 T}{r_{1}} + P_{atm}$
For bubble 2,
$P_{2} - P_{atm} = \frac{4 T}{r_{2}}$
$P_{2} = \frac{4 T}{r_{2}} + P_{atm}$
Consider that two bubbles are merged,
$P_{1} - P_{2} = \frac{4 T}{r}$
$\frac{4 T}{r_{1}} + P_{atm} - \frac{4 T}{r_{2}} - P_{atm} = \frac{4 T}{r}$
$(\frac{1}{r_{1}} - \frac{1}{r_{2}}) = \frac{1}{r}$
$r = \frac{r_{1} r_{2}}{r_{2} - r_{1}}$

JCECE-2004

Mechanical Properties of Fluids

143030 The amount of work done in blowing a soap bubble such that its diameter increases from $d$ to $D$ is (T-surface tension of the solution)

1

4 π (D^{2} - d^{2}) T

2

8 π (D^{2} - d^{2}) T

3

π (D^{2} - d^{2}) T

4

2 π (D^{2} - d^{2}) T

Explanation:

D We know that,
$New surface area = π D^{2}$
$Old surface area = π d^{2}$
Increase in area of soap bubble $(Δ A) = 2 (π D^{2} - π d^{2})$
$= 2 π (D^{2} - d^{2})$
Work done $(W) =$ Increase in area of soap bubble $\times$ Surface tension
$W = 2 π (D^{2} - d^{2}) T$

CG PET- 2004

Mechanical Properties of Fluids

143031 Consider an air bubble of radius $2 mm$ in a liquid at a depth of $5 cm$ below the free surface. The density of the liquid is $1000 kg / m^{3}$ and the surface tension is $0.1 N / m$ . Then, find the pressure inside the air bubble is greater than the pressure at the free surface of the liquid. (Take, $g = 10 m / s^{2}$ )

1

500 Pa

2

600 Pa

3

700 Pa

4

800 Pa

Explanation:

B Given,
Radius of bubble $(r) = 2 mm = 0.002 m$
Depth of bubble below the water surface $(h) = 5 cm$ $= 0.05 m$
Density of liquid $(ρ) = 1000 kg / m^{3}$
Surface tension of liquid $(σ) = 0.1 N / m$
$g = 10 m / \sec^{2}$
Gauge pressure due to liquid as a depth of $5 cm -$
$P = ρ gh$
$P = 1000 \times 10 \times 0.05$
$P = 500 Pa$
Excess pressure inside the air bubble w.r.t. outside due to surface tension -
$P^{'} = \frac{2 σ}{r}$
$P^{'} = \frac{2 \times 0.1}{0.002}$
$P^{'} = 100 Pa$
Total excess pressure inside the bubble w.r.t. to surface liquid -
$P_{total} = P + P^{'}$
$P_{total} = 500 + 100$
$P_{total} = 600 Pa$

TS- EAMCET-07.05.2018

Mechanical Properties of Fluids

143032 Consider a water droplet of diameter $0.2 mm$ where the outside pressure is $1.5 N / {cm}^{2}$ at $25^{\circ} C$ . the pressure inside the droplet, when the surface tension at $25^{\circ} C$ is $0.08 N / m$ is

1

0.32 N / {cm}^{2}

2

1.18 N / {cm}^{2}

3

1.82 N / {cm}^{2}

4

1.66 N / {cm}^{2}

Explanation:

D Given,
Diameter of the water droplet $(d) = 0.2 mm$
Radius of the water droplet $r = 0.1 mm = 0.0001 m$
Pressure outside droplet $(P_{0}) = 1.5 N / {cm}^{2}$
$= 1.5 \times 10^{4} N / m^{2}$
Surface tension $(σ) = 0.08 N / m$ We know,
Pressure inside droplet $(P) = \frac{2 σ}{r} + P_{0}$
$P = \frac{2 \times 0.08}{0.0001} + 1.5 \times 10^{4}$
$P = 0.16 \times 10^{4} + 1.5 \times 10^{4}$
$P = 1.66 N / {cm}^{2}$

TS- EAMCET-06.05.2019