QBManager

Mechanical Properties of Fluids

142820 A U-shaped wire is dipped in a soap solution and removed. The thin soap film formed between the wire and light slider supports a weight of $1.5 \times 10^{2} N$ . The length of the slider is $30 cm$ . What is the surface tension of the film?

1

3 \times 10^{- 3} {Nm}^{- 1}

2

2 \times 10^{- 5} {Nm}^{- 1}

3

4 \times 10^{- 4} {Nm}^{- 1}

4

2.5 \times 10^{- 2} {Nm}^{- 1}

Explanation:

D Given, $w = 1.5 \times 10^{2} N, l = 30 cm$
We know that,
Surface tension, $T = \frac{F}{l}$
Total length of the film Suspension,
$(l) = 2 \times 30 cm = 0.60 m$
$F = mg$
$T = \frac{mg}{l}$
$T = \frac{1.5 \times 10^{- 2}}{0.6}$
$T = 2.5 \times 10^{- 2} N / m$
$T = 2.5 \times 10^{- 2} {Nm}^{- 1}$

BCECE-2015

Mechanical Properties of Fluids

142821 The surface tension of soap solution is $σ$ . What is the work done in blowing soap bubble of radius $r$ ?

1

π r^{2} σ

2

2 π r^{2} σ

3

4 π r^{2} σ

4

8 π r^{2} σ

Explanation:

D Work done $=$ Surface tension $\times 2 (Δ A)$
Change in Area $(Δ A) = 2 \times 4 π (R_{1}^{2} - R_{2}^{2})$
$= 2 \times 4 π r^{2} [∴ R_{1}^{2} - R_{2}^{2} = r^{2}]$
$Δ A = 8 π r^{2}$
Work done $= σ \times Δ A$
$= σ \times 8 π r^{2}$
$= 8 σ π r^{2}$

BCECE-2009

Mechanical Properties of Fluids

142822 A soap film of surface tension $3 \times 10^{- 2} {Nm}^{- 1}$ formed in rectangular frame, can support a straw. If the length of the film is $10 cm$ , then the mass of the straw that film can support is-

1

0.06 g

2

0.6 g

3

6 g

4

60 g

Explanation:

B Given that,
$T = 3 \times 10^{- 2} N / m, l = 10 cm$
We know that,
$T = \frac{F}{l}$
$F = T \times l ($ For rectangular frame $l = 2 l)$
$mg = 2 T l$
$m = \frac{2 T l}{g}$
$m = \frac{2 \times (3 \times 10^{- 2}) \times 10 \times 10^{- 2}}{9.8} kg$
$m = 6.1 \times 10^{- 4} kg ≃ 0.6 g$

BCECE-2012

Mechanical Properties of Fluids

142823 The surface tension of soap solution is $0.03 N / m$ . The amount of work done in forming a bubble of radius $5 cm$ is-

1

3.77 J

2

1.885 J

3

0.95 \times 10^{- 3} J

4

1.9 \times 10^{- 3} J

Explanation:

D Given that,
$T = 0.03 N / m$
$r = 5 cm = 5 \times 10^{- 2}$
We know that,
Work done for bubble $(W) = T \times 2 (A)$
$= T \times 2 (4 π r^{2})$
$= 8 π r^{2} T$
$W = 8 \times 3.14 \times {(5 \times 10^{- 2})}^{2} \times 0.03$
$W = 1.88 \times 10^{- 3} = 1.9 \times 10^{- 3} J$

BCECE-2011

Mechanical Properties of Fluids

142824 The material of a wire has a density of 1.4 $g / {cm}^{3}$ . It is not wetted by a liquid of surface tension $44 dyne / cm$ , then the maximum radius of the wire which can float on the surface of liquid is-

1

\frac{10}{28} cm

2

\frac{10}{14} cm

3

\frac{10}{7} mm

4

0.7 cm

Explanation:

C We know that,
Weight of the wire $=$ Force due to surface tension
$mg = T \times (2 l)$
$(ρ V) g = 2 T l$
$ρ (π r^{2} l) g = 2 T l$
$r = \sqrt{(\frac{2 T}{π ρ g})}$
$= \sqrt{\frac{2 \times 44 \times 7}{22 \times 1.4 \times 980}}$
$= \frac{1}{7} cm$
$= \frac{10}{7} mm$

BCECE-2011

Mechanical Properties of Fluids

142820 A U-shaped wire is dipped in a soap solution and removed. The thin soap film formed between the wire and light slider supports a weight of $1.5 \times 10^{2} N$ . The length of the slider is $30 cm$ . What is the surface tension of the film?

1

3 \times 10^{- 3} {Nm}^{- 1}

2

2 \times 10^{- 5} {Nm}^{- 1}

3

4 \times 10^{- 4} {Nm}^{- 1}

4

2.5 \times 10^{- 2} {Nm}^{- 1}

Explanation:

D Given, $w = 1.5 \times 10^{2} N, l = 30 cm$
We know that,
Surface tension, $T = \frac{F}{l}$
Total length of the film Suspension,
$(l) = 2 \times 30 cm = 0.60 m$
$F = mg$
$T = \frac{mg}{l}$
$T = \frac{1.5 \times 10^{- 2}}{0.6}$
$T = 2.5 \times 10^{- 2} N / m$
$T = 2.5 \times 10^{- 2} {Nm}^{- 1}$

BCECE-2015

Mechanical Properties of Fluids

142821 The surface tension of soap solution is $σ$ . What is the work done in blowing soap bubble of radius $r$ ?

1

π r^{2} σ

2

2 π r^{2} σ

3

4 π r^{2} σ

4

8 π r^{2} σ

Explanation:

D Work done $=$ Surface tension $\times 2 (Δ A)$
Change in Area $(Δ A) = 2 \times 4 π (R_{1}^{2} - R_{2}^{2})$
$= 2 \times 4 π r^{2} [∴ R_{1}^{2} - R_{2}^{2} = r^{2}]$
$Δ A = 8 π r^{2}$
Work done $= σ \times Δ A$
$= σ \times 8 π r^{2}$
$= 8 σ π r^{2}$

BCECE-2009

Mechanical Properties of Fluids

142822 A soap film of surface tension $3 \times 10^{- 2} {Nm}^{- 1}$ formed in rectangular frame, can support a straw. If the length of the film is $10 cm$ , then the mass of the straw that film can support is-

1

0.06 g

2

0.6 g

3

6 g

4

60 g

Explanation:

B Given that,
$T = 3 \times 10^{- 2} N / m, l = 10 cm$
We know that,
$T = \frac{F}{l}$
$F = T \times l ($ For rectangular frame $l = 2 l)$
$mg = 2 T l$
$m = \frac{2 T l}{g}$
$m = \frac{2 \times (3 \times 10^{- 2}) \times 10 \times 10^{- 2}}{9.8} kg$
$m = 6.1 \times 10^{- 4} kg ≃ 0.6 g$

BCECE-2012

Mechanical Properties of Fluids

142823 The surface tension of soap solution is $0.03 N / m$ . The amount of work done in forming a bubble of radius $5 cm$ is-

1

3.77 J

2

1.885 J

3

0.95 \times 10^{- 3} J

4

1.9 \times 10^{- 3} J

Explanation:

D Given that,
$T = 0.03 N / m$
$r = 5 cm = 5 \times 10^{- 2}$
We know that,
Work done for bubble $(W) = T \times 2 (A)$
$= T \times 2 (4 π r^{2})$
$= 8 π r^{2} T$
$W = 8 \times 3.14 \times {(5 \times 10^{- 2})}^{2} \times 0.03$
$W = 1.88 \times 10^{- 3} = 1.9 \times 10^{- 3} J$

BCECE-2011

Mechanical Properties of Fluids

142824 The material of a wire has a density of 1.4 $g / {cm}^{3}$ . It is not wetted by a liquid of surface tension $44 dyne / cm$ , then the maximum radius of the wire which can float on the surface of liquid is-

1

\frac{10}{28} cm

2

\frac{10}{14} cm

3

\frac{10}{7} mm

4

0.7 cm

Explanation:

C We know that,
Weight of the wire $=$ Force due to surface tension
$mg = T \times (2 l)$
$(ρ V) g = 2 T l$
$ρ (π r^{2} l) g = 2 T l$
$r = \sqrt{(\frac{2 T}{π ρ g})}$
$= \sqrt{\frac{2 \times 44 \times 7}{22 \times 1.4 \times 980}}$
$= \frac{1}{7} cm$
$= \frac{10}{7} mm$

BCECE-2011

Mechanical Properties of Fluids

142820 A U-shaped wire is dipped in a soap solution and removed. The thin soap film formed between the wire and light slider supports a weight of $1.5 \times 10^{2} N$ . The length of the slider is $30 cm$ . What is the surface tension of the film?

1

3 \times 10^{- 3} {Nm}^{- 1}

2

2 \times 10^{- 5} {Nm}^{- 1}

3

4 \times 10^{- 4} {Nm}^{- 1}

4

2.5 \times 10^{- 2} {Nm}^{- 1}

Explanation:

D Given, $w = 1.5 \times 10^{2} N, l = 30 cm$
We know that,
Surface tension, $T = \frac{F}{l}$
Total length of the film Suspension,
$(l) = 2 \times 30 cm = 0.60 m$
$F = mg$
$T = \frac{mg}{l}$
$T = \frac{1.5 \times 10^{- 2}}{0.6}$
$T = 2.5 \times 10^{- 2} N / m$
$T = 2.5 \times 10^{- 2} {Nm}^{- 1}$

BCECE-2015

Mechanical Properties of Fluids

142821 The surface tension of soap solution is $σ$ . What is the work done in blowing soap bubble of radius $r$ ?

1

π r^{2} σ

2

2 π r^{2} σ

3

4 π r^{2} σ

4

8 π r^{2} σ

Explanation:

D Work done $=$ Surface tension $\times 2 (Δ A)$
Change in Area $(Δ A) = 2 \times 4 π (R_{1}^{2} - R_{2}^{2})$
$= 2 \times 4 π r^{2} [∴ R_{1}^{2} - R_{2}^{2} = r^{2}]$
$Δ A = 8 π r^{2}$
Work done $= σ \times Δ A$
$= σ \times 8 π r^{2}$
$= 8 σ π r^{2}$

BCECE-2009

Mechanical Properties of Fluids

142822 A soap film of surface tension $3 \times 10^{- 2} {Nm}^{- 1}$ formed in rectangular frame, can support a straw. If the length of the film is $10 cm$ , then the mass of the straw that film can support is-

1

0.06 g

2

0.6 g

3

6 g

4

60 g

Explanation:

B Given that,
$T = 3 \times 10^{- 2} N / m, l = 10 cm$
We know that,
$T = \frac{F}{l}$
$F = T \times l ($ For rectangular frame $l = 2 l)$
$mg = 2 T l$
$m = \frac{2 T l}{g}$
$m = \frac{2 \times (3 \times 10^{- 2}) \times 10 \times 10^{- 2}}{9.8} kg$
$m = 6.1 \times 10^{- 4} kg ≃ 0.6 g$

BCECE-2012

Mechanical Properties of Fluids

142823 The surface tension of soap solution is $0.03 N / m$ . The amount of work done in forming a bubble of radius $5 cm$ is-

1

3.77 J

2

1.885 J

3

0.95 \times 10^{- 3} J

4

1.9 \times 10^{- 3} J

Explanation:

D Given that,
$T = 0.03 N / m$
$r = 5 cm = 5 \times 10^{- 2}$
We know that,
Work done for bubble $(W) = T \times 2 (A)$
$= T \times 2 (4 π r^{2})$
$= 8 π r^{2} T$
$W = 8 \times 3.14 \times {(5 \times 10^{- 2})}^{2} \times 0.03$
$W = 1.88 \times 10^{- 3} = 1.9 \times 10^{- 3} J$

BCECE-2011

Mechanical Properties of Fluids

142824 The material of a wire has a density of 1.4 $g / {cm}^{3}$ . It is not wetted by a liquid of surface tension $44 dyne / cm$ , then the maximum radius of the wire which can float on the surface of liquid is-

1

\frac{10}{28} cm

2

\frac{10}{14} cm

3

\frac{10}{7} mm

4

0.7 cm

Explanation:

C We know that,
Weight of the wire $=$ Force due to surface tension
$mg = T \times (2 l)$
$(ρ V) g = 2 T l$
$ρ (π r^{2} l) g = 2 T l$
$r = \sqrt{(\frac{2 T}{π ρ g})}$
$= \sqrt{\frac{2 \times 44 \times 7}{22 \times 1.4 \times 980}}$
$= \frac{1}{7} cm$
$= \frac{10}{7} mm$

BCECE-2011

Mechanical Properties of Fluids

142820 A U-shaped wire is dipped in a soap solution and removed. The thin soap film formed between the wire and light slider supports a weight of $1.5 \times 10^{2} N$ . The length of the slider is $30 cm$ . What is the surface tension of the film?

1

3 \times 10^{- 3} {Nm}^{- 1}

2

2 \times 10^{- 5} {Nm}^{- 1}

3

4 \times 10^{- 4} {Nm}^{- 1}

4

2.5 \times 10^{- 2} {Nm}^{- 1}

Explanation:

D Given, $w = 1.5 \times 10^{2} N, l = 30 cm$
We know that,
Surface tension, $T = \frac{F}{l}$
Total length of the film Suspension,
$(l) = 2 \times 30 cm = 0.60 m$
$F = mg$
$T = \frac{mg}{l}$
$T = \frac{1.5 \times 10^{- 2}}{0.6}$
$T = 2.5 \times 10^{- 2} N / m$
$T = 2.5 \times 10^{- 2} {Nm}^{- 1}$

BCECE-2015

Mechanical Properties of Fluids

142821 The surface tension of soap solution is $σ$ . What is the work done in blowing soap bubble of radius $r$ ?

1

π r^{2} σ

2

2 π r^{2} σ

3

4 π r^{2} σ

4

8 π r^{2} σ

Explanation:

D Work done $=$ Surface tension $\times 2 (Δ A)$
Change in Area $(Δ A) = 2 \times 4 π (R_{1}^{2} - R_{2}^{2})$
$= 2 \times 4 π r^{2} [∴ R_{1}^{2} - R_{2}^{2} = r^{2}]$
$Δ A = 8 π r^{2}$
Work done $= σ \times Δ A$
$= σ \times 8 π r^{2}$
$= 8 σ π r^{2}$

BCECE-2009

Mechanical Properties of Fluids

142822 A soap film of surface tension $3 \times 10^{- 2} {Nm}^{- 1}$ formed in rectangular frame, can support a straw. If the length of the film is $10 cm$ , then the mass of the straw that film can support is-

1

0.06 g

2

0.6 g

3

6 g

4

60 g

Explanation:

B Given that,
$T = 3 \times 10^{- 2} N / m, l = 10 cm$
We know that,
$T = \frac{F}{l}$
$F = T \times l ($ For rectangular frame $l = 2 l)$
$mg = 2 T l$
$m = \frac{2 T l}{g}$
$m = \frac{2 \times (3 \times 10^{- 2}) \times 10 \times 10^{- 2}}{9.8} kg$
$m = 6.1 \times 10^{- 4} kg ≃ 0.6 g$

BCECE-2012

Mechanical Properties of Fluids

142823 The surface tension of soap solution is $0.03 N / m$ . The amount of work done in forming a bubble of radius $5 cm$ is-

1

3.77 J

2

1.885 J

3

0.95 \times 10^{- 3} J

4

1.9 \times 10^{- 3} J

Explanation:

D Given that,
$T = 0.03 N / m$
$r = 5 cm = 5 \times 10^{- 2}$
We know that,
Work done for bubble $(W) = T \times 2 (A)$
$= T \times 2 (4 π r^{2})$
$= 8 π r^{2} T$
$W = 8 \times 3.14 \times {(5 \times 10^{- 2})}^{2} \times 0.03$
$W = 1.88 \times 10^{- 3} = 1.9 \times 10^{- 3} J$

BCECE-2011

Mechanical Properties of Fluids

142824 The material of a wire has a density of 1.4 $g / {cm}^{3}$ . It is not wetted by a liquid of surface tension $44 dyne / cm$ , then the maximum radius of the wire which can float on the surface of liquid is-

1

\frac{10}{28} cm

2

\frac{10}{14} cm

3

\frac{10}{7} mm

4

0.7 cm

Explanation:

C We know that,
Weight of the wire $=$ Force due to surface tension
$mg = T \times (2 l)$
$(ρ V) g = 2 T l$
$ρ (π r^{2} l) g = 2 T l$
$r = \sqrt{(\frac{2 T}{π ρ g})}$
$= \sqrt{\frac{2 \times 44 \times 7}{22 \times 1.4 \times 980}}$
$= \frac{1}{7} cm$
$= \frac{10}{7} mm$

BCECE-2011

Mechanical Properties of Fluids

142820 A U-shaped wire is dipped in a soap solution and removed. The thin soap film formed between the wire and light slider supports a weight of $1.5 \times 10^{2} N$ . The length of the slider is $30 cm$ . What is the surface tension of the film?

1

3 \times 10^{- 3} {Nm}^{- 1}

2

2 \times 10^{- 5} {Nm}^{- 1}

3

4 \times 10^{- 4} {Nm}^{- 1}

4

2.5 \times 10^{- 2} {Nm}^{- 1}

Explanation:

D Given, $w = 1.5 \times 10^{2} N, l = 30 cm$
We know that,
Surface tension, $T = \frac{F}{l}$
Total length of the film Suspension,
$(l) = 2 \times 30 cm = 0.60 m$
$F = mg$
$T = \frac{mg}{l}$
$T = \frac{1.5 \times 10^{- 2}}{0.6}$
$T = 2.5 \times 10^{- 2} N / m$
$T = 2.5 \times 10^{- 2} {Nm}^{- 1}$

BCECE-2015

Mechanical Properties of Fluids

142821 The surface tension of soap solution is $σ$ . What is the work done in blowing soap bubble of radius $r$ ?

1

π r^{2} σ

2

2 π r^{2} σ

3

4 π r^{2} σ

4

8 π r^{2} σ

Explanation:

D Work done $=$ Surface tension $\times 2 (Δ A)$
Change in Area $(Δ A) = 2 \times 4 π (R_{1}^{2} - R_{2}^{2})$
$= 2 \times 4 π r^{2} [∴ R_{1}^{2} - R_{2}^{2} = r^{2}]$
$Δ A = 8 π r^{2}$
Work done $= σ \times Δ A$
$= σ \times 8 π r^{2}$
$= 8 σ π r^{2}$

BCECE-2009

Mechanical Properties of Fluids

142822 A soap film of surface tension $3 \times 10^{- 2} {Nm}^{- 1}$ formed in rectangular frame, can support a straw. If the length of the film is $10 cm$ , then the mass of the straw that film can support is-

1

0.06 g

2

0.6 g

3

6 g

4

60 g

Explanation:

B Given that,
$T = 3 \times 10^{- 2} N / m, l = 10 cm$
We know that,
$T = \frac{F}{l}$
$F = T \times l ($ For rectangular frame $l = 2 l)$
$mg = 2 T l$
$m = \frac{2 T l}{g}$
$m = \frac{2 \times (3 \times 10^{- 2}) \times 10 \times 10^{- 2}}{9.8} kg$
$m = 6.1 \times 10^{- 4} kg ≃ 0.6 g$

BCECE-2012

Mechanical Properties of Fluids

142823 The surface tension of soap solution is $0.03 N / m$ . The amount of work done in forming a bubble of radius $5 cm$ is-

1

3.77 J

2

1.885 J

3

0.95 \times 10^{- 3} J

4

1.9 \times 10^{- 3} J

Explanation:

D Given that,
$T = 0.03 N / m$
$r = 5 cm = 5 \times 10^{- 2}$
We know that,
Work done for bubble $(W) = T \times 2 (A)$
$= T \times 2 (4 π r^{2})$
$= 8 π r^{2} T$
$W = 8 \times 3.14 \times {(5 \times 10^{- 2})}^{2} \times 0.03$
$W = 1.88 \times 10^{- 3} = 1.9 \times 10^{- 3} J$

BCECE-2011

Mechanical Properties of Fluids

142824 The material of a wire has a density of 1.4 $g / {cm}^{3}$ . It is not wetted by a liquid of surface tension $44 dyne / cm$ , then the maximum radius of the wire which can float on the surface of liquid is-

1

\frac{10}{28} cm

2

\frac{10}{14} cm

3

\frac{10}{7} mm

4

0.7 cm

Explanation:

C We know that,
Weight of the wire $=$ Force due to surface tension
$mg = T \times (2 l)$
$(ρ V) g = 2 T l$
$ρ (π r^{2} l) g = 2 T l$
$r = \sqrt{(\frac{2 T}{π ρ g})}$
$= \sqrt{\frac{2 \times 44 \times 7}{22 \times 1.4 \times 980}}$
$= \frac{1}{7} cm$
$= \frac{10}{7} mm$

BCECE-2011

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