QBManager

Mechanical Properties of Fluids

142787 A spherical body of density $ρ$ is floating half immersed in a liquid of density $d$ . If $σ$ is the surface tension of the liquid, then the diameter of the body is

1

\sqrt{\frac{3 σ}{g (2 ρ - d)}}

2

\sqrt{\frac{6 σ}{g (2 ρ - d)}}

3

\sqrt{\frac{4 σ}{g (2 ρ - d)}}

4

\sqrt{\frac{12 σ}{g (2 ρ - d)}}

Explanation:

D
Balancing the surface tension of the drop
$=$ buoyant force + force of surface tension
$\Rightarrow \frac{4}{3} π r^{3} ρ g = 2 π r σ + \frac{1}{2} (\frac{4}{3} π r^{3} dg)$
$\Rightarrow 2 π r σ = \frac{π r^{3} g}{3} [4 ρ - 2 d]$
$\Rightarrow r^{2} = \frac{3 \times 2 π σ}{π g (4 ρ - 2 d)}$
$\Rightarrow r^{2} = \frac{3 σ}{g (2 ρ - d)}$
$\Rightarrow r = \sqrt{\frac{3 σ}{g (2 ρ - d)}}$
$Diameter = 2 r = \sqrt{\frac{12 σ}{g (2 ρ - d)}}$

AP EAMCET (22.04.2018) Shift-II

Mechanical Properties of Fluids

142788 Two circular plates of radius $5 cm$ each, have a $0.01 mm$ thick water film between them. Then what will be the force required to separate these plate $($ S.T. of water $= 73 dyne / cm)$ ?

1

125 N

2

95 N

3

115 N

4

105 N

Explanation:

C Given, surface tension of water $= 73$ dyne $/ cm$ We know that,
$W = F \times d$
$W = T \times A \times 2$
by using equation (i) and (ii), we get
$F \times d = T \times A \times 2$
$F = \frac{2 \times T \times A}{d}$
$F = \frac{2 \times T \times π r^{2}}{d}$
$F = \frac{2 \times 73 \times 10^{- 3} \times π \times (0.05)^{2}}{0.01 \times 10^{- 3}}$
$F = 36.5 π ≃ 115$ Newton

BITSAT-2010

Mechanical Properties of Fluids

142789 If the masses of all molecules of a gas are halved and their speeds doubled, then the ratio of initial and final pressure will be

1

1 : 4

2

4 : 1

3

2 : 1

4

1 : 2

Explanation:

D According to the kinetic energy gas equation $P = \frac{1}{3} \frac{{mNv}^{2}}{V}$
$∴ P_{i} = \frac{{Nmv}^{2}}{3 V}$
and $P_{f} = \frac{N (\frac{1}{2} m) (2 v)^{2}}{3 V} = \frac{2 {Nmv}^{2}}{3 V}$ ${$ Given, $v_{f} = 2 v, m_{f} = m / 2}$
Ratio of pressure from initial to final
$∴ \frac{P_{i}}{P_{f}} = \frac{\frac{{Nmv}^{2}}{3 V}}{\frac{2 {Nmv}^{2}}{3 V}} = \frac{1}{2}$
Hence, ratio of initial and final pressure is $1 : 2$ .

CG PET 2019

Mechanical Properties of Fluids

142790 Drop of liquid of density $ρ$ is floating half immersed in a liquid of density $ρ_{0}$ . If surface tension of liquid is $s$ , the radius of drop is

1

\sqrt{\frac{3 s}{g (ρ - ρ_{0})}}

2

\sqrt{\frac{3 s}{g (2 ρ - ρ_{0})}}

3

\sqrt{\frac{3 s}{g (3 ρ - ρ_{0})}}

4

\sqrt{\frac{3 s}{g (4 ρ - ρ_{0})}}

Explanation:

B Given,

Density of drop $= ρ$
Density of liquid $= ρ_{0}$
Surface tension $= S$
Volume of drop $= v$
Balance force in y direction
$F_{G} = F_{B} + F^{'}$
$mg = (\frac{v}{2} ρ_{0}) g + s \times 2 π r$
$vdg = \frac{v}{2} ρ_{0} g + s \times 2 π r$
$\frac{4}{3} π r^{3} ρ g = \frac{4}{6} π r^{3} ρ_{0} g + s \times 2 π r$
$\frac{4}{3} π r^{3} g (ρ - \frac{ρ_{0}}{2}) = s \times 2 π r$
$r^{2} = \frac{3 s / 2}{g (ρ - \frac{ρ_{0}}{2})} = \frac{3 s}{g (2 ρ - ρ_{0})}$
$r = \sqrt{\frac{3 s}{g (2 ρ - ρ_{0})}}$

CG PET 2019

Mechanical Properties of Fluids

142787 A spherical body of density $ρ$ is floating half immersed in a liquid of density $d$ . If $σ$ is the surface tension of the liquid, then the diameter of the body is

1

\sqrt{\frac{3 σ}{g (2 ρ - d)}}

2

\sqrt{\frac{6 σ}{g (2 ρ - d)}}

3

\sqrt{\frac{4 σ}{g (2 ρ - d)}}

4

\sqrt{\frac{12 σ}{g (2 ρ - d)}}

Explanation:

D
Balancing the surface tension of the drop
$=$ buoyant force + force of surface tension
$\Rightarrow \frac{4}{3} π r^{3} ρ g = 2 π r σ + \frac{1}{2} (\frac{4}{3} π r^{3} dg)$
$\Rightarrow 2 π r σ = \frac{π r^{3} g}{3} [4 ρ - 2 d]$
$\Rightarrow r^{2} = \frac{3 \times 2 π σ}{π g (4 ρ - 2 d)}$
$\Rightarrow r^{2} = \frac{3 σ}{g (2 ρ - d)}$
$\Rightarrow r = \sqrt{\frac{3 σ}{g (2 ρ - d)}}$
$Diameter = 2 r = \sqrt{\frac{12 σ}{g (2 ρ - d)}}$

AP EAMCET (22.04.2018) Shift-II

Mechanical Properties of Fluids

142788 Two circular plates of radius $5 cm$ each, have a $0.01 mm$ thick water film between them. Then what will be the force required to separate these plate $($ S.T. of water $= 73 dyne / cm)$ ?

1

125 N

2

95 N

3

115 N

4

105 N

Explanation:

C Given, surface tension of water $= 73$ dyne $/ cm$ We know that,
$W = F \times d$
$W = T \times A \times 2$
by using equation (i) and (ii), we get
$F \times d = T \times A \times 2$
$F = \frac{2 \times T \times A}{d}$
$F = \frac{2 \times T \times π r^{2}}{d}$
$F = \frac{2 \times 73 \times 10^{- 3} \times π \times (0.05)^{2}}{0.01 \times 10^{- 3}}$
$F = 36.5 π ≃ 115$ Newton

BITSAT-2010

Mechanical Properties of Fluids

142789 If the masses of all molecules of a gas are halved and their speeds doubled, then the ratio of initial and final pressure will be

1

1 : 4

2

4 : 1

3

2 : 1

4

1 : 2

Explanation:

D According to the kinetic energy gas equation $P = \frac{1}{3} \frac{{mNv}^{2}}{V}$
$∴ P_{i} = \frac{{Nmv}^{2}}{3 V}$
and $P_{f} = \frac{N (\frac{1}{2} m) (2 v)^{2}}{3 V} = \frac{2 {Nmv}^{2}}{3 V}$ ${$ Given, $v_{f} = 2 v, m_{f} = m / 2}$
Ratio of pressure from initial to final
$∴ \frac{P_{i}}{P_{f}} = \frac{\frac{{Nmv}^{2}}{3 V}}{\frac{2 {Nmv}^{2}}{3 V}} = \frac{1}{2}$
Hence, ratio of initial and final pressure is $1 : 2$ .

CG PET 2019

Mechanical Properties of Fluids

142790 Drop of liquid of density $ρ$ is floating half immersed in a liquid of density $ρ_{0}$ . If surface tension of liquid is $s$ , the radius of drop is

1

\sqrt{\frac{3 s}{g (ρ - ρ_{0})}}

2

\sqrt{\frac{3 s}{g (2 ρ - ρ_{0})}}

3

\sqrt{\frac{3 s}{g (3 ρ - ρ_{0})}}

4

\sqrt{\frac{3 s}{g (4 ρ - ρ_{0})}}

Explanation:

B Given,

Density of drop $= ρ$
Density of liquid $= ρ_{0}$
Surface tension $= S$
Volume of drop $= v$
Balance force in y direction
$F_{G} = F_{B} + F^{'}$
$mg = (\frac{v}{2} ρ_{0}) g + s \times 2 π r$
$vdg = \frac{v}{2} ρ_{0} g + s \times 2 π r$
$\frac{4}{3} π r^{3} ρ g = \frac{4}{6} π r^{3} ρ_{0} g + s \times 2 π r$
$\frac{4}{3} π r^{3} g (ρ - \frac{ρ_{0}}{2}) = s \times 2 π r$
$r^{2} = \frac{3 s / 2}{g (ρ - \frac{ρ_{0}}{2})} = \frac{3 s}{g (2 ρ - ρ_{0})}$
$r = \sqrt{\frac{3 s}{g (2 ρ - ρ_{0})}}$

CG PET 2019

Mechanical Properties of Fluids

142787 A spherical body of density $ρ$ is floating half immersed in a liquid of density $d$ . If $σ$ is the surface tension of the liquid, then the diameter of the body is

1

\sqrt{\frac{3 σ}{g (2 ρ - d)}}

2

\sqrt{\frac{6 σ}{g (2 ρ - d)}}

3

\sqrt{\frac{4 σ}{g (2 ρ - d)}}

4

\sqrt{\frac{12 σ}{g (2 ρ - d)}}

Explanation:

D
Balancing the surface tension of the drop
$=$ buoyant force + force of surface tension
$\Rightarrow \frac{4}{3} π r^{3} ρ g = 2 π r σ + \frac{1}{2} (\frac{4}{3} π r^{3} dg)$
$\Rightarrow 2 π r σ = \frac{π r^{3} g}{3} [4 ρ - 2 d]$
$\Rightarrow r^{2} = \frac{3 \times 2 π σ}{π g (4 ρ - 2 d)}$
$\Rightarrow r^{2} = \frac{3 σ}{g (2 ρ - d)}$
$\Rightarrow r = \sqrt{\frac{3 σ}{g (2 ρ - d)}}$
$Diameter = 2 r = \sqrt{\frac{12 σ}{g (2 ρ - d)}}$

AP EAMCET (22.04.2018) Shift-II

Mechanical Properties of Fluids

142788 Two circular plates of radius $5 cm$ each, have a $0.01 mm$ thick water film between them. Then what will be the force required to separate these plate $($ S.T. of water $= 73 dyne / cm)$ ?

1

125 N

2

95 N

3

115 N

4

105 N

Explanation:

C Given, surface tension of water $= 73$ dyne $/ cm$ We know that,
$W = F \times d$
$W = T \times A \times 2$
by using equation (i) and (ii), we get
$F \times d = T \times A \times 2$
$F = \frac{2 \times T \times A}{d}$
$F = \frac{2 \times T \times π r^{2}}{d}$
$F = \frac{2 \times 73 \times 10^{- 3} \times π \times (0.05)^{2}}{0.01 \times 10^{- 3}}$
$F = 36.5 π ≃ 115$ Newton

BITSAT-2010

Mechanical Properties of Fluids

142789 If the masses of all molecules of a gas are halved and their speeds doubled, then the ratio of initial and final pressure will be

1

1 : 4

2

4 : 1

3

2 : 1

4

1 : 2

Explanation:

D According to the kinetic energy gas equation $P = \frac{1}{3} \frac{{mNv}^{2}}{V}$
$∴ P_{i} = \frac{{Nmv}^{2}}{3 V}$
and $P_{f} = \frac{N (\frac{1}{2} m) (2 v)^{2}}{3 V} = \frac{2 {Nmv}^{2}}{3 V}$ ${$ Given, $v_{f} = 2 v, m_{f} = m / 2}$
Ratio of pressure from initial to final
$∴ \frac{P_{i}}{P_{f}} = \frac{\frac{{Nmv}^{2}}{3 V}}{\frac{2 {Nmv}^{2}}{3 V}} = \frac{1}{2}$
Hence, ratio of initial and final pressure is $1 : 2$ .

CG PET 2019

Mechanical Properties of Fluids

142790 Drop of liquid of density $ρ$ is floating half immersed in a liquid of density $ρ_{0}$ . If surface tension of liquid is $s$ , the radius of drop is

1

\sqrt{\frac{3 s}{g (ρ - ρ_{0})}}

2

\sqrt{\frac{3 s}{g (2 ρ - ρ_{0})}}

3

\sqrt{\frac{3 s}{g (3 ρ - ρ_{0})}}

4

\sqrt{\frac{3 s}{g (4 ρ - ρ_{0})}}

Explanation:

B Given,

Density of drop $= ρ$
Density of liquid $= ρ_{0}$
Surface tension $= S$
Volume of drop $= v$
Balance force in y direction
$F_{G} = F_{B} + F^{'}$
$mg = (\frac{v}{2} ρ_{0}) g + s \times 2 π r$
$vdg = \frac{v}{2} ρ_{0} g + s \times 2 π r$
$\frac{4}{3} π r^{3} ρ g = \frac{4}{6} π r^{3} ρ_{0} g + s \times 2 π r$
$\frac{4}{3} π r^{3} g (ρ - \frac{ρ_{0}}{2}) = s \times 2 π r$
$r^{2} = \frac{3 s / 2}{g (ρ - \frac{ρ_{0}}{2})} = \frac{3 s}{g (2 ρ - ρ_{0})}$
$r = \sqrt{\frac{3 s}{g (2 ρ - ρ_{0})}}$

CG PET 2019

Mechanical Properties of Fluids

142787 A spherical body of density $ρ$ is floating half immersed in a liquid of density $d$ . If $σ$ is the surface tension of the liquid, then the diameter of the body is

1

\sqrt{\frac{3 σ}{g (2 ρ - d)}}

2

\sqrt{\frac{6 σ}{g (2 ρ - d)}}

3

\sqrt{\frac{4 σ}{g (2 ρ - d)}}

4

\sqrt{\frac{12 σ}{g (2 ρ - d)}}

Explanation:

D
Balancing the surface tension of the drop
$=$ buoyant force + force of surface tension
$\Rightarrow \frac{4}{3} π r^{3} ρ g = 2 π r σ + \frac{1}{2} (\frac{4}{3} π r^{3} dg)$
$\Rightarrow 2 π r σ = \frac{π r^{3} g}{3} [4 ρ - 2 d]$
$\Rightarrow r^{2} = \frac{3 \times 2 π σ}{π g (4 ρ - 2 d)}$
$\Rightarrow r^{2} = \frac{3 σ}{g (2 ρ - d)}$
$\Rightarrow r = \sqrt{\frac{3 σ}{g (2 ρ - d)}}$
$Diameter = 2 r = \sqrt{\frac{12 σ}{g (2 ρ - d)}}$

AP EAMCET (22.04.2018) Shift-II

Mechanical Properties of Fluids

142788 Two circular plates of radius $5 cm$ each, have a $0.01 mm$ thick water film between them. Then what will be the force required to separate these plate $($ S.T. of water $= 73 dyne / cm)$ ?

1

125 N

2

95 N

3

115 N

4

105 N

Explanation:

C Given, surface tension of water $= 73$ dyne $/ cm$ We know that,
$W = F \times d$
$W = T \times A \times 2$
by using equation (i) and (ii), we get
$F \times d = T \times A \times 2$
$F = \frac{2 \times T \times A}{d}$
$F = \frac{2 \times T \times π r^{2}}{d}$
$F = \frac{2 \times 73 \times 10^{- 3} \times π \times (0.05)^{2}}{0.01 \times 10^{- 3}}$
$F = 36.5 π ≃ 115$ Newton

BITSAT-2010

Mechanical Properties of Fluids

142789 If the masses of all molecules of a gas are halved and their speeds doubled, then the ratio of initial and final pressure will be

1

1 : 4

2

4 : 1

3

2 : 1

4

1 : 2

Explanation:

D According to the kinetic energy gas equation $P = \frac{1}{3} \frac{{mNv}^{2}}{V}$
$∴ P_{i} = \frac{{Nmv}^{2}}{3 V}$
and $P_{f} = \frac{N (\frac{1}{2} m) (2 v)^{2}}{3 V} = \frac{2 {Nmv}^{2}}{3 V}$ ${$ Given, $v_{f} = 2 v, m_{f} = m / 2}$
Ratio of pressure from initial to final
$∴ \frac{P_{i}}{P_{f}} = \frac{\frac{{Nmv}^{2}}{3 V}}{\frac{2 {Nmv}^{2}}{3 V}} = \frac{1}{2}$
Hence, ratio of initial and final pressure is $1 : 2$ .

CG PET 2019

Mechanical Properties of Fluids

142790 Drop of liquid of density $ρ$ is floating half immersed in a liquid of density $ρ_{0}$ . If surface tension of liquid is $s$ , the radius of drop is

1

\sqrt{\frac{3 s}{g (ρ - ρ_{0})}}

2

\sqrt{\frac{3 s}{g (2 ρ - ρ_{0})}}

3

\sqrt{\frac{3 s}{g (3 ρ - ρ_{0})}}

4

\sqrt{\frac{3 s}{g (4 ρ - ρ_{0})}}

Explanation:

B Given,

Density of drop $= ρ$
Density of liquid $= ρ_{0}$
Surface tension $= S$
Volume of drop $= v$
Balance force in y direction
$F_{G} = F_{B} + F^{'}$
$mg = (\frac{v}{2} ρ_{0}) g + s \times 2 π r$
$vdg = \frac{v}{2} ρ_{0} g + s \times 2 π r$
$\frac{4}{3} π r^{3} ρ g = \frac{4}{6} π r^{3} ρ_{0} g + s \times 2 π r$
$\frac{4}{3} π r^{3} g (ρ - \frac{ρ_{0}}{2}) = s \times 2 π r$
$r^{2} = \frac{3 s / 2}{g (ρ - \frac{ρ_{0}}{2})} = \frac{3 s}{g (2 ρ - ρ_{0})}$
$r = \sqrt{\frac{3 s}{g (2 ρ - ρ_{0})}}$

CG PET 2019