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Mechanical Properties of Fluids

143101 In a cylinder provided with a pistion, air is under pressure $P_{1}$ at a constant temperature ' $t$ '. $A$ soap bubble with radius ' $r$ ' and surface tension ' $T$ ' is lying inside the cylinder. To reduce the radius of the soap bubble to half, the required air pressure inside the cylinder is

1

8 P_{1} + \frac{24 T}{r}

2

8 P_{1} + \frac{3 T}{r}

3

8 P_{1} + \frac{2 T}{r}

4

8 P_{1} + \frac{12 T}{r}

Explanation:

A Given,
Initial pressure of a soap bubble $= P_{1}$
Initial radius of a soap bubble $= r$
Final pressure after compression $= P_{2}$
Surface tension of soap bubble $= T$
Final radius of a soap bubble $= r / 2$
Now, Temperature remains constant
$(P_{1} V_{1} = P_{2} V_{2}$
$∴ \frac{4 T}{r}) (\frac{4}{3} π r^{3}) = (P_{2} + \frac{4 T}{r / 2}) (\frac{4}{3} π \frac{r^{3}}{8})$
$8 P_{1} + \frac{32 T}{r} = P_{2} + \frac{8 T}{r}$
$\Rightarrow P_{2} = 8 P_{1} + \frac{24 T}{r}$

AP EAMCET-25.04.2017

Mechanical Properties of Fluids

143102 A soap bubble has radius ' $R$ ' and thickness $d (<< R)$ . It collapses into a spherical drop. The ratio of excess pressure in the drop to the excess pressure inside the bubble is

1

{(\frac{R}{3 d})}^{\frac{1}{3}}

2

{(\frac{R}{8 d})}^{\frac{1}{3}}

3

{(\frac{R}{6 d})}^{\frac{1}{3}}

4

{(\frac{R}{24 d})}^{\frac{1}{3}}

Explanation:

D Given that,
Radius of soap bubble $= R$
Thickness of bubble layer $= d$
Pressure, $P_{1} = \frac{4 T}{R}, P_{2} = \frac{2 T}{r}$

$R ≫ d$
Where $P_{1}$ is excess pressure of soap bubble.
Let $r$ is the radius of spherical drop
$\frac{P_{2}}{P_{1}} = \frac{R}{2 r}$
$∵ 4 π R^{2} d = \frac{4}{3} π r^{3}$
$r = {(3 R^{2} d)}^{1 / 3}$
Putting the value of $r$ in equation (i),
$\frac{P_{2}}{P_{1}} = \frac{R}{2 {(3 R^{2} d)}^{1 / 3}}$
$\frac{P_{2}}{P_{1}} = \frac{R}{{(24 R^{2} d)}^{1 / 3}}$
$\frac{P_{2}}{P_{1}} = {(\frac{R}{24 d})}^{1 / 3}$

AP EAMCET-28.04.2017

Mechanical Properties of Fluids

143103 In S.I. system, the total energy of the free surface of a liquid drop is $2 π$ times the surface tension of the liquid .The diameter of the drop is

1

\sqrt{3} m

2

\sqrt{2} m

3

\sqrt{5} m

4

\sqrt{7} m

Explanation:

B Given that,
Total energy of liquid drop $(E) = 2 π T$
$E = T . Δ A$
$2 π T = T .4 π r^{2} [∵ A = 4 π r^{2}]$
$2 π T = 4 T π r^{2}$
$2 r^{2} = 1$
$r^{2} = \frac{1}{2}$
$r = \frac{1}{\sqrt{2}}$
Diameter $(d) = 2 r$
$d = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2}$
$d = \sqrt{2}$
The diameter of the drop is $\sqrt{2}$ .

AP EAMCET-26.04.2017

Mechanical Properties of Fluids

143104 A $31.4 kg$ girl wearing high heel shoes balances her weight on a single heel. The diameter of the heel is $1 cm$ . The pressure (in pascal) exerted by the heel on the floor is (Take $g = 10 m / s^{2}$ )

1

2 \times 10^{6}

2

4 \times 10^{6}

3

6 \times 10^{6}

4

8 \times 10^{6}

Explanation:

B Given that,
Mass of the girl $(m) = 31.4 kg$
Diameter of the heal $= 1 cm = 0.01 m$
Radius of the heal $= 0.005 m$
Gravitational Acceleration $(g) = 10 m / s^{2}$
Area of the heal $= π r^{2}$
$= π (0.005)^{2}$
$= 3.14 \times 25 \times 10^{- 6}$
$= 7.85 \times 10^{- 5} m^{2}$
Force exerted by the heal on the floor,
$F = mg$
$F = 31.4 \times 10$
$F = 314 N$
Pressure exerted by the heal on the floor
$P = \frac{F}{A}$
$Pressure (P) = \frac{314}{7.85 \times 10^{- 5}}$
$= \frac{314 \times 10^{5}}{7.85}$
$= 40 \times 10^{5}$
$= 4 \times 10^{6} N / m^{2}$
The pressure exerted by the heal on the floor is 4 $\times 10^{6} N / m^{2}$

AMU-2019

Mechanical Properties of Fluids

143105 A soap bubble of radius $r$ is blown further to increase its radius to $2 r$ under isothermal conditions. If $σ$ is the surface tension of the soap solution, the energy spent in doing so is

1

4 π σ r^{2}

2

8 π σ r^{2}

3

12 π σ r^{2}

4

24 π σ r^{2}

Explanation:

D Given,
$r_{1} = r$
$r_{2} = 2 r$
Surface tension $= σ$
Surface area of bubble of Radius (r) $= 4 π r^{2}$
New surface area $= 16 π r^{2}$
Increases surface area $(Δ A) = 16 π r^{2} - 4 π r^{2}$
$Δ A = 12 π r^{2}$
A bubble has two surface, the total increase in surface area $= 2 \times A = 2 \times 12 π r^{2}$
$= 24 π r^{2}$
Energy spent $=$ Work done $= σ \times$ Area $= σ \times 24 π r^{2}$
$= 24 π σ r^{2}$

AP EAMCET(Medical)-2014

Mechanical Properties of Fluids

143101 In a cylinder provided with a pistion, air is under pressure $P_{1}$ at a constant temperature ' $t$ '. $A$ soap bubble with radius ' $r$ ' and surface tension ' $T$ ' is lying inside the cylinder. To reduce the radius of the soap bubble to half, the required air pressure inside the cylinder is

1

8 P_{1} + \frac{24 T}{r}

2

8 P_{1} + \frac{3 T}{r}

3

8 P_{1} + \frac{2 T}{r}

4

8 P_{1} + \frac{12 T}{r}

Explanation:

A Given,
Initial pressure of a soap bubble $= P_{1}$
Initial radius of a soap bubble $= r$
Final pressure after compression $= P_{2}$
Surface tension of soap bubble $= T$
Final radius of a soap bubble $= r / 2$
Now, Temperature remains constant
$(P_{1} V_{1} = P_{2} V_{2}$
$∴ \frac{4 T}{r}) (\frac{4}{3} π r^{3}) = (P_{2} + \frac{4 T}{r / 2}) (\frac{4}{3} π \frac{r^{3}}{8})$
$8 P_{1} + \frac{32 T}{r} = P_{2} + \frac{8 T}{r}$
$\Rightarrow P_{2} = 8 P_{1} + \frac{24 T}{r}$

AP EAMCET-25.04.2017

Mechanical Properties of Fluids

143102 A soap bubble has radius ' $R$ ' and thickness $d (<< R)$ . It collapses into a spherical drop. The ratio of excess pressure in the drop to the excess pressure inside the bubble is

1

{(\frac{R}{3 d})}^{\frac{1}{3}}

2

{(\frac{R}{8 d})}^{\frac{1}{3}}

3

{(\frac{R}{6 d})}^{\frac{1}{3}}

4

{(\frac{R}{24 d})}^{\frac{1}{3}}

Explanation:

D Given that,
Radius of soap bubble $= R$
Thickness of bubble layer $= d$
Pressure, $P_{1} = \frac{4 T}{R}, P_{2} = \frac{2 T}{r}$

$R ≫ d$
Where $P_{1}$ is excess pressure of soap bubble.
Let $r$ is the radius of spherical drop
$\frac{P_{2}}{P_{1}} = \frac{R}{2 r}$
$∵ 4 π R^{2} d = \frac{4}{3} π r^{3}$
$r = {(3 R^{2} d)}^{1 / 3}$
Putting the value of $r$ in equation (i),
$\frac{P_{2}}{P_{1}} = \frac{R}{2 {(3 R^{2} d)}^{1 / 3}}$
$\frac{P_{2}}{P_{1}} = \frac{R}{{(24 R^{2} d)}^{1 / 3}}$
$\frac{P_{2}}{P_{1}} = {(\frac{R}{24 d})}^{1 / 3}$

AP EAMCET-28.04.2017

Mechanical Properties of Fluids

143103 In S.I. system, the total energy of the free surface of a liquid drop is $2 π$ times the surface tension of the liquid .The diameter of the drop is

1

\sqrt{3} m

2

\sqrt{2} m

3

\sqrt{5} m

4

\sqrt{7} m

Explanation:

B Given that,
Total energy of liquid drop $(E) = 2 π T$
$E = T . Δ A$
$2 π T = T .4 π r^{2} [∵ A = 4 π r^{2}]$
$2 π T = 4 T π r^{2}$
$2 r^{2} = 1$
$r^{2} = \frac{1}{2}$
$r = \frac{1}{\sqrt{2}}$
Diameter $(d) = 2 r$
$d = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2}$
$d = \sqrt{2}$
The diameter of the drop is $\sqrt{2}$ .

AP EAMCET-26.04.2017

Mechanical Properties of Fluids

143104 A $31.4 kg$ girl wearing high heel shoes balances her weight on a single heel. The diameter of the heel is $1 cm$ . The pressure (in pascal) exerted by the heel on the floor is (Take $g = 10 m / s^{2}$ )

1

2 \times 10^{6}

2

4 \times 10^{6}

3

6 \times 10^{6}

4

8 \times 10^{6}

Explanation:

B Given that,
Mass of the girl $(m) = 31.4 kg$
Diameter of the heal $= 1 cm = 0.01 m$
Radius of the heal $= 0.005 m$
Gravitational Acceleration $(g) = 10 m / s^{2}$
Area of the heal $= π r^{2}$
$= π (0.005)^{2}$
$= 3.14 \times 25 \times 10^{- 6}$
$= 7.85 \times 10^{- 5} m^{2}$
Force exerted by the heal on the floor,
$F = mg$
$F = 31.4 \times 10$
$F = 314 N$
Pressure exerted by the heal on the floor
$P = \frac{F}{A}$
$Pressure (P) = \frac{314}{7.85 \times 10^{- 5}}$
$= \frac{314 \times 10^{5}}{7.85}$
$= 40 \times 10^{5}$
$= 4 \times 10^{6} N / m^{2}$
The pressure exerted by the heal on the floor is 4 $\times 10^{6} N / m^{2}$

AMU-2019

Mechanical Properties of Fluids

143105 A soap bubble of radius $r$ is blown further to increase its radius to $2 r$ under isothermal conditions. If $σ$ is the surface tension of the soap solution, the energy spent in doing so is

1

4 π σ r^{2}

2

8 π σ r^{2}

3

12 π σ r^{2}

4

24 π σ r^{2}

Explanation:

D Given,
$r_{1} = r$
$r_{2} = 2 r$
Surface tension $= σ$
Surface area of bubble of Radius (r) $= 4 π r^{2}$
New surface area $= 16 π r^{2}$
Increases surface area $(Δ A) = 16 π r^{2} - 4 π r^{2}$
$Δ A = 12 π r^{2}$
A bubble has two surface, the total increase in surface area $= 2 \times A = 2 \times 12 π r^{2}$
$= 24 π r^{2}$
Energy spent $=$ Work done $= σ \times$ Area $= σ \times 24 π r^{2}$
$= 24 π σ r^{2}$

AP EAMCET(Medical)-2014

Mechanical Properties of Fluids

143101 In a cylinder provided with a pistion, air is under pressure $P_{1}$ at a constant temperature ' $t$ '. $A$ soap bubble with radius ' $r$ ' and surface tension ' $T$ ' is lying inside the cylinder. To reduce the radius of the soap bubble to half, the required air pressure inside the cylinder is

1

8 P_{1} + \frac{24 T}{r}

2

8 P_{1} + \frac{3 T}{r}

3

8 P_{1} + \frac{2 T}{r}

4

8 P_{1} + \frac{12 T}{r}

Explanation:

A Given,
Initial pressure of a soap bubble $= P_{1}$
Initial radius of a soap bubble $= r$
Final pressure after compression $= P_{2}$
Surface tension of soap bubble $= T$
Final radius of a soap bubble $= r / 2$
Now, Temperature remains constant
$(P_{1} V_{1} = P_{2} V_{2}$
$∴ \frac{4 T}{r}) (\frac{4}{3} π r^{3}) = (P_{2} + \frac{4 T}{r / 2}) (\frac{4}{3} π \frac{r^{3}}{8})$
$8 P_{1} + \frac{32 T}{r} = P_{2} + \frac{8 T}{r}$
$\Rightarrow P_{2} = 8 P_{1} + \frac{24 T}{r}$

AP EAMCET-25.04.2017

Mechanical Properties of Fluids

143102 A soap bubble has radius ' $R$ ' and thickness $d (<< R)$ . It collapses into a spherical drop. The ratio of excess pressure in the drop to the excess pressure inside the bubble is

1

{(\frac{R}{3 d})}^{\frac{1}{3}}

2

{(\frac{R}{8 d})}^{\frac{1}{3}}

3

{(\frac{R}{6 d})}^{\frac{1}{3}}

4

{(\frac{R}{24 d})}^{\frac{1}{3}}

Explanation:

D Given that,
Radius of soap bubble $= R$
Thickness of bubble layer $= d$
Pressure, $P_{1} = \frac{4 T}{R}, P_{2} = \frac{2 T}{r}$

$R ≫ d$
Where $P_{1}$ is excess pressure of soap bubble.
Let $r$ is the radius of spherical drop
$\frac{P_{2}}{P_{1}} = \frac{R}{2 r}$
$∵ 4 π R^{2} d = \frac{4}{3} π r^{3}$
$r = {(3 R^{2} d)}^{1 / 3}$
Putting the value of $r$ in equation (i),
$\frac{P_{2}}{P_{1}} = \frac{R}{2 {(3 R^{2} d)}^{1 / 3}}$
$\frac{P_{2}}{P_{1}} = \frac{R}{{(24 R^{2} d)}^{1 / 3}}$
$\frac{P_{2}}{P_{1}} = {(\frac{R}{24 d})}^{1 / 3}$

AP EAMCET-28.04.2017

Mechanical Properties of Fluids

143103 In S.I. system, the total energy of the free surface of a liquid drop is $2 π$ times the surface tension of the liquid .The diameter of the drop is

1

\sqrt{3} m

2

\sqrt{2} m

3

\sqrt{5} m

4

\sqrt{7} m

Explanation:

B Given that,
Total energy of liquid drop $(E) = 2 π T$
$E = T . Δ A$
$2 π T = T .4 π r^{2} [∵ A = 4 π r^{2}]$
$2 π T = 4 T π r^{2}$
$2 r^{2} = 1$
$r^{2} = \frac{1}{2}$
$r = \frac{1}{\sqrt{2}}$
Diameter $(d) = 2 r$
$d = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2}$
$d = \sqrt{2}$
The diameter of the drop is $\sqrt{2}$ .

AP EAMCET-26.04.2017

Mechanical Properties of Fluids

143104 A $31.4 kg$ girl wearing high heel shoes balances her weight on a single heel. The diameter of the heel is $1 cm$ . The pressure (in pascal) exerted by the heel on the floor is (Take $g = 10 m / s^{2}$ )

1

2 \times 10^{6}

2

4 \times 10^{6}

3

6 \times 10^{6}

4

8 \times 10^{6}

Explanation:

B Given that,
Mass of the girl $(m) = 31.4 kg$
Diameter of the heal $= 1 cm = 0.01 m$
Radius of the heal $= 0.005 m$
Gravitational Acceleration $(g) = 10 m / s^{2}$
Area of the heal $= π r^{2}$
$= π (0.005)^{2}$
$= 3.14 \times 25 \times 10^{- 6}$
$= 7.85 \times 10^{- 5} m^{2}$
Force exerted by the heal on the floor,
$F = mg$
$F = 31.4 \times 10$
$F = 314 N$
Pressure exerted by the heal on the floor
$P = \frac{F}{A}$
$Pressure (P) = \frac{314}{7.85 \times 10^{- 5}}$
$= \frac{314 \times 10^{5}}{7.85}$
$= 40 \times 10^{5}$
$= 4 \times 10^{6} N / m^{2}$
The pressure exerted by the heal on the floor is 4 $\times 10^{6} N / m^{2}$

AMU-2019

Mechanical Properties of Fluids

143105 A soap bubble of radius $r$ is blown further to increase its radius to $2 r$ under isothermal conditions. If $σ$ is the surface tension of the soap solution, the energy spent in doing so is

1

4 π σ r^{2}

2

8 π σ r^{2}

3

12 π σ r^{2}

4

24 π σ r^{2}

Explanation:

D Given,
$r_{1} = r$
$r_{2} = 2 r$
Surface tension $= σ$
Surface area of bubble of Radius (r) $= 4 π r^{2}$
New surface area $= 16 π r^{2}$
Increases surface area $(Δ A) = 16 π r^{2} - 4 π r^{2}$
$Δ A = 12 π r^{2}$
A bubble has two surface, the total increase in surface area $= 2 \times A = 2 \times 12 π r^{2}$
$= 24 π r^{2}$
Energy spent $=$ Work done $= σ \times$ Area $= σ \times 24 π r^{2}$
$= 24 π σ r^{2}$

AP EAMCET(Medical)-2014

Mechanical Properties of Fluids

143101 In a cylinder provided with a pistion, air is under pressure $P_{1}$ at a constant temperature ' $t$ '. $A$ soap bubble with radius ' $r$ ' and surface tension ' $T$ ' is lying inside the cylinder. To reduce the radius of the soap bubble to half, the required air pressure inside the cylinder is

1

8 P_{1} + \frac{24 T}{r}

2

8 P_{1} + \frac{3 T}{r}

3

8 P_{1} + \frac{2 T}{r}

4

8 P_{1} + \frac{12 T}{r}

Explanation:

A Given,
Initial pressure of a soap bubble $= P_{1}$
Initial radius of a soap bubble $= r$
Final pressure after compression $= P_{2}$
Surface tension of soap bubble $= T$
Final radius of a soap bubble $= r / 2$
Now, Temperature remains constant
$(P_{1} V_{1} = P_{2} V_{2}$
$∴ \frac{4 T}{r}) (\frac{4}{3} π r^{3}) = (P_{2} + \frac{4 T}{r / 2}) (\frac{4}{3} π \frac{r^{3}}{8})$
$8 P_{1} + \frac{32 T}{r} = P_{2} + \frac{8 T}{r}$
$\Rightarrow P_{2} = 8 P_{1} + \frac{24 T}{r}$

AP EAMCET-25.04.2017

Mechanical Properties of Fluids

143102 A soap bubble has radius ' $R$ ' and thickness $d (<< R)$ . It collapses into a spherical drop. The ratio of excess pressure in the drop to the excess pressure inside the bubble is

1

{(\frac{R}{3 d})}^{\frac{1}{3}}

2

{(\frac{R}{8 d})}^{\frac{1}{3}}

3

{(\frac{R}{6 d})}^{\frac{1}{3}}

4

{(\frac{R}{24 d})}^{\frac{1}{3}}

Explanation:

D Given that,
Radius of soap bubble $= R$
Thickness of bubble layer $= d$
Pressure, $P_{1} = \frac{4 T}{R}, P_{2} = \frac{2 T}{r}$

$R ≫ d$
Where $P_{1}$ is excess pressure of soap bubble.
Let $r$ is the radius of spherical drop
$\frac{P_{2}}{P_{1}} = \frac{R}{2 r}$
$∵ 4 π R^{2} d = \frac{4}{3} π r^{3}$
$r = {(3 R^{2} d)}^{1 / 3}$
Putting the value of $r$ in equation (i),
$\frac{P_{2}}{P_{1}} = \frac{R}{2 {(3 R^{2} d)}^{1 / 3}}$
$\frac{P_{2}}{P_{1}} = \frac{R}{{(24 R^{2} d)}^{1 / 3}}$
$\frac{P_{2}}{P_{1}} = {(\frac{R}{24 d})}^{1 / 3}$

AP EAMCET-28.04.2017

Mechanical Properties of Fluids

143103 In S.I. system, the total energy of the free surface of a liquid drop is $2 π$ times the surface tension of the liquid .The diameter of the drop is

1

\sqrt{3} m

2

\sqrt{2} m

3

\sqrt{5} m

4

\sqrt{7} m

Explanation:

B Given that,
Total energy of liquid drop $(E) = 2 π T$
$E = T . Δ A$
$2 π T = T .4 π r^{2} [∵ A = 4 π r^{2}]$
$2 π T = 4 T π r^{2}$
$2 r^{2} = 1$
$r^{2} = \frac{1}{2}$
$r = \frac{1}{\sqrt{2}}$
Diameter $(d) = 2 r$
$d = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2}$
$d = \sqrt{2}$
The diameter of the drop is $\sqrt{2}$ .

AP EAMCET-26.04.2017

Mechanical Properties of Fluids

143104 A $31.4 kg$ girl wearing high heel shoes balances her weight on a single heel. The diameter of the heel is $1 cm$ . The pressure (in pascal) exerted by the heel on the floor is (Take $g = 10 m / s^{2}$ )

1

2 \times 10^{6}

2

4 \times 10^{6}

3

6 \times 10^{6}

4

8 \times 10^{6}

Explanation:

B Given that,
Mass of the girl $(m) = 31.4 kg$
Diameter of the heal $= 1 cm = 0.01 m$
Radius of the heal $= 0.005 m$
Gravitational Acceleration $(g) = 10 m / s^{2}$
Area of the heal $= π r^{2}$
$= π (0.005)^{2}$
$= 3.14 \times 25 \times 10^{- 6}$
$= 7.85 \times 10^{- 5} m^{2}$
Force exerted by the heal on the floor,
$F = mg$
$F = 31.4 \times 10$
$F = 314 N$
Pressure exerted by the heal on the floor
$P = \frac{F}{A}$
$Pressure (P) = \frac{314}{7.85 \times 10^{- 5}}$
$= \frac{314 \times 10^{5}}{7.85}$
$= 40 \times 10^{5}$
$= 4 \times 10^{6} N / m^{2}$
The pressure exerted by the heal on the floor is 4 $\times 10^{6} N / m^{2}$

AMU-2019

Mechanical Properties of Fluids

143105 A soap bubble of radius $r$ is blown further to increase its radius to $2 r$ under isothermal conditions. If $σ$ is the surface tension of the soap solution, the energy spent in doing so is

1

4 π σ r^{2}

2

8 π σ r^{2}

3

12 π σ r^{2}

4

24 π σ r^{2}

Explanation:

D Given,
$r_{1} = r$
$r_{2} = 2 r$
Surface tension $= σ$
Surface area of bubble of Radius (r) $= 4 π r^{2}$
New surface area $= 16 π r^{2}$
Increases surface area $(Δ A) = 16 π r^{2} - 4 π r^{2}$
$Δ A = 12 π r^{2}$
A bubble has two surface, the total increase in surface area $= 2 \times A = 2 \times 12 π r^{2}$
$= 24 π r^{2}$
Energy spent $=$ Work done $= σ \times$ Area $= σ \times 24 π r^{2}$
$= 24 π σ r^{2}$

AP EAMCET(Medical)-2014

Mechanical Properties of Fluids

143101 In a cylinder provided with a pistion, air is under pressure $P_{1}$ at a constant temperature ' $t$ '. $A$ soap bubble with radius ' $r$ ' and surface tension ' $T$ ' is lying inside the cylinder. To reduce the radius of the soap bubble to half, the required air pressure inside the cylinder is

1

8 P_{1} + \frac{24 T}{r}

2

8 P_{1} + \frac{3 T}{r}

3

8 P_{1} + \frac{2 T}{r}

4

8 P_{1} + \frac{12 T}{r}

Explanation:

A Given,
Initial pressure of a soap bubble $= P_{1}$
Initial radius of a soap bubble $= r$
Final pressure after compression $= P_{2}$
Surface tension of soap bubble $= T$
Final radius of a soap bubble $= r / 2$
Now, Temperature remains constant
$(P_{1} V_{1} = P_{2} V_{2}$
$∴ \frac{4 T}{r}) (\frac{4}{3} π r^{3}) = (P_{2} + \frac{4 T}{r / 2}) (\frac{4}{3} π \frac{r^{3}}{8})$
$8 P_{1} + \frac{32 T}{r} = P_{2} + \frac{8 T}{r}$
$\Rightarrow P_{2} = 8 P_{1} + \frac{24 T}{r}$

AP EAMCET-25.04.2017

Mechanical Properties of Fluids

143102 A soap bubble has radius ' $R$ ' and thickness $d (<< R)$ . It collapses into a spherical drop. The ratio of excess pressure in the drop to the excess pressure inside the bubble is

1

{(\frac{R}{3 d})}^{\frac{1}{3}}

2

{(\frac{R}{8 d})}^{\frac{1}{3}}

3

{(\frac{R}{6 d})}^{\frac{1}{3}}

4

{(\frac{R}{24 d})}^{\frac{1}{3}}

Explanation:

D Given that,
Radius of soap bubble $= R$
Thickness of bubble layer $= d$
Pressure, $P_{1} = \frac{4 T}{R}, P_{2} = \frac{2 T}{r}$

$R ≫ d$
Where $P_{1}$ is excess pressure of soap bubble.
Let $r$ is the radius of spherical drop
$\frac{P_{2}}{P_{1}} = \frac{R}{2 r}$
$∵ 4 π R^{2} d = \frac{4}{3} π r^{3}$
$r = {(3 R^{2} d)}^{1 / 3}$
Putting the value of $r$ in equation (i),
$\frac{P_{2}}{P_{1}} = \frac{R}{2 {(3 R^{2} d)}^{1 / 3}}$
$\frac{P_{2}}{P_{1}} = \frac{R}{{(24 R^{2} d)}^{1 / 3}}$
$\frac{P_{2}}{P_{1}} = {(\frac{R}{24 d})}^{1 / 3}$

AP EAMCET-28.04.2017

Mechanical Properties of Fluids

143103 In S.I. system, the total energy of the free surface of a liquid drop is $2 π$ times the surface tension of the liquid .The diameter of the drop is

1

\sqrt{3} m

2

\sqrt{2} m

3

\sqrt{5} m

4

\sqrt{7} m

Explanation:

B Given that,
Total energy of liquid drop $(E) = 2 π T$
$E = T . Δ A$
$2 π T = T .4 π r^{2} [∵ A = 4 π r^{2}]$
$2 π T = 4 T π r^{2}$
$2 r^{2} = 1$
$r^{2} = \frac{1}{2}$
$r = \frac{1}{\sqrt{2}}$
Diameter $(d) = 2 r$
$d = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2}$
$d = \sqrt{2}$
The diameter of the drop is $\sqrt{2}$ .

AP EAMCET-26.04.2017

Mechanical Properties of Fluids

143104 A $31.4 kg$ girl wearing high heel shoes balances her weight on a single heel. The diameter of the heel is $1 cm$ . The pressure (in pascal) exerted by the heel on the floor is (Take $g = 10 m / s^{2}$ )

1

2 \times 10^{6}

2

4 \times 10^{6}

3

6 \times 10^{6}

4

8 \times 10^{6}

Explanation:

B Given that,
Mass of the girl $(m) = 31.4 kg$
Diameter of the heal $= 1 cm = 0.01 m$
Radius of the heal $= 0.005 m$
Gravitational Acceleration $(g) = 10 m / s^{2}$
Area of the heal $= π r^{2}$
$= π (0.005)^{2}$
$= 3.14 \times 25 \times 10^{- 6}$
$= 7.85 \times 10^{- 5} m^{2}$
Force exerted by the heal on the floor,
$F = mg$
$F = 31.4 \times 10$
$F = 314 N$
Pressure exerted by the heal on the floor
$P = \frac{F}{A}$
$Pressure (P) = \frac{314}{7.85 \times 10^{- 5}}$
$= \frac{314 \times 10^{5}}{7.85}$
$= 40 \times 10^{5}$
$= 4 \times 10^{6} N / m^{2}$
The pressure exerted by the heal on the floor is 4 $\times 10^{6} N / m^{2}$

AMU-2019

Mechanical Properties of Fluids

143105 A soap bubble of radius $r$ is blown further to increase its radius to $2 r$ under isothermal conditions. If $σ$ is the surface tension of the soap solution, the energy spent in doing so is

1

4 π σ r^{2}

2

8 π σ r^{2}

3

12 π σ r^{2}

4

24 π σ r^{2}

Explanation:

D Given,
$r_{1} = r$
$r_{2} = 2 r$
Surface tension $= σ$
Surface area of bubble of Radius (r) $= 4 π r^{2}$
New surface area $= 16 π r^{2}$
Increases surface area $(Δ A) = 16 π r^{2} - 4 π r^{2}$
$Δ A = 12 π r^{2}$
A bubble has two surface, the total increase in surface area $= 2 \times A = 2 \times 12 π r^{2}$
$= 24 π r^{2}$
Energy spent $=$ Work done $= σ \times$ Area $= σ \times 24 π r^{2}$
$= 24 π σ r^{2}$

AP EAMCET(Medical)-2014