QBManager

Mechanical Properties of Fluids

142825 The surface tension of a liquid is $5 N / m$ . If a film is held on a ring of area $0.02 m^{2}$ , its total surface energy is about :

1

2 \times 10^{- 2} J

2

2.5 \times 10^{- 2} J

3

2 \times 10^{- 1} J

4

3 \times 10^{- 1} J

Explanation:

C Given that,
$T = 5 N / m$
$A = 0.02 m^{2}$
We know that,
Work done or surface energy of the film $(E) = 2 T$ A
$= 2 \times 5 \times 0.02$
$= 2 \times 5 \times 0.02$
$= 2 \times 10^{- 1} J$

BCECE-2005

Mechanical Properties of Fluids

142826 Pressure inside two soap bubbles are 1.01 and 1.02 atm. Ratio between their volumes is :

1

102 : 101

2

(102)^{3} : (103)^{3}

3

8 : 1

4

2 : 1

Explanation:

C Given that,
Pressure inside first bubble $= 1.01 atm$
Pressure inside second bubble $= 1.02 atm$
Since atmospheric pressure $= 1 atm$ So,
Excess pressure $Δ P_{1} = 1.01 - 1 = 0.01 atm$
Excess pressure $Δ P_{2} = 1.02 - 1 = 0.02 atm$
Change in pressure,
$Δ P = \frac{4 T}{r}$
$∴ Δ P \propto \frac{1}{r}$
$\frac{Δ P_{2}}{Δ P_{1}} = \frac{r_{1}}{r_{2}}$
$\frac{0.02}{0.01} = \frac{r_{1}}{r_{2}}$
$\frac{r_{1}}{r_{2}} = \frac{2}{1}$
Now,
$V = \frac{4}{3} π r^{3}$
$∴ \frac{V_{1}}{V_{2}} = {(\frac{r_{1}}{r_{2}})}^{3} = {(\frac{2}{1})}^{3} = \frac{8}{1}$
$V_{1} : V_{2} = 8 : 1$

BCECE-2003

Mechanical Properties of Fluids

142827 A water film is formed between the two straight parallel wires, each of length $10 cm$ , kept at a separation of $0.5 cm$ . Now, the separation between them is increased by $1 mm$ without breaking the water film. The work done for this is
( Surface tension of water $= 7.2 \times 10^{- 2} {Nm}^{- 1}$ )

1

1.44 \times 10^{- 5} J

2

5.76 \times 10^{- 5} J

3

7.22 \times 10^{- 6} J

4

2.88 \times 10^{- 5} J

Explanation:

A Given that,
Length of wire, $l = 10 cm$
Separation between them, $b_{1} = 0.5 cm$
If the separation increased by $= 1 mm = 0.1 cm$
$b_{2} = b_{1} + 0.5 = 0.1 + 0.5 = 0.6 cm$
$Δ A = (A_{2} - A_{1})$
$Δ A = 2 l (b_{2} - b_{1}) [∵ 2$ Surfaces $]$
$Δ A = 2 \times 10 (0.6 - 0.5)$
$Δ A = 2 \times 10^{- 4} m^{2}$
We know that,
$∵$ Surface tension $(T) = 7.2 \times 10^{- 2} N / m$
$∴$ Work done $(W) = T \times Δ A$
$W = 7.2 \times 10^{- 2} \times 2 \times 10^{- 4}$
$W = 14.4 \times 10^{- 6}$
$W = 1.44 \times 10^{- 5} N . m$

MHT-CET 2020

Mechanical Properties of Fluids

142829 If the surface tension of a soap solution is $3 \times$ $10^{- 2} N / m$ , then the work done in forming a soap film of $20 cm \times 5 cm$ will be

1

6 \times 10^{- 3} J

2

6 J

3

6 \times 10^{- 4} J

4

6 \times 10^{- 2} J

Explanation:

C Given that,
Total surface area in which soap film is formed $A = 2 \times 20 \times 5$ [Soap bubble form 2 surface Area]
$A = 200 {cm}^{2}$
$A = 200 \times 10^{- 4} m^{2}$
$A = 2 \times 10^{- 2} m^{2}$
Surface tension $(T) = 3 \times 10^{- 2} N / m$
Work done $(W) = T \times A$
$W = 3 \times 10^{- 2} \times 2 \times 10^{- 2}$
$W = 6 \times 10^{- 4} J$

MHT-CET 2020

Mechanical Properties of Fluids

142825 The surface tension of a liquid is $5 N / m$ . If a film is held on a ring of area $0.02 m^{2}$ , its total surface energy is about :

1

2 \times 10^{- 2} J

2

2.5 \times 10^{- 2} J

3

2 \times 10^{- 1} J

4

3 \times 10^{- 1} J

Explanation:

C Given that,
$T = 5 N / m$
$A = 0.02 m^{2}$
We know that,
Work done or surface energy of the film $(E) = 2 T$ A
$= 2 \times 5 \times 0.02$
$= 2 \times 5 \times 0.02$
$= 2 \times 10^{- 1} J$

BCECE-2005

Mechanical Properties of Fluids

142826 Pressure inside two soap bubbles are 1.01 and 1.02 atm. Ratio between their volumes is :

1

102 : 101

2

(102)^{3} : (103)^{3}

3

8 : 1

4

2 : 1

Explanation:

C Given that,
Pressure inside first bubble $= 1.01 atm$
Pressure inside second bubble $= 1.02 atm$
Since atmospheric pressure $= 1 atm$ So,
Excess pressure $Δ P_{1} = 1.01 - 1 = 0.01 atm$
Excess pressure $Δ P_{2} = 1.02 - 1 = 0.02 atm$
Change in pressure,
$Δ P = \frac{4 T}{r}$
$∴ Δ P \propto \frac{1}{r}$
$\frac{Δ P_{2}}{Δ P_{1}} = \frac{r_{1}}{r_{2}}$
$\frac{0.02}{0.01} = \frac{r_{1}}{r_{2}}$
$\frac{r_{1}}{r_{2}} = \frac{2}{1}$
Now,
$V = \frac{4}{3} π r^{3}$
$∴ \frac{V_{1}}{V_{2}} = {(\frac{r_{1}}{r_{2}})}^{3} = {(\frac{2}{1})}^{3} = \frac{8}{1}$
$V_{1} : V_{2} = 8 : 1$

BCECE-2003

Mechanical Properties of Fluids

142827 A water film is formed between the two straight parallel wires, each of length $10 cm$ , kept at a separation of $0.5 cm$ . Now, the separation between them is increased by $1 mm$ without breaking the water film. The work done for this is
( Surface tension of water $= 7.2 \times 10^{- 2} {Nm}^{- 1}$ )

1

1.44 \times 10^{- 5} J

2

5.76 \times 10^{- 5} J

3

7.22 \times 10^{- 6} J

4

2.88 \times 10^{- 5} J

Explanation:

A Given that,
Length of wire, $l = 10 cm$
Separation between them, $b_{1} = 0.5 cm$
If the separation increased by $= 1 mm = 0.1 cm$
$b_{2} = b_{1} + 0.5 = 0.1 + 0.5 = 0.6 cm$
$Δ A = (A_{2} - A_{1})$
$Δ A = 2 l (b_{2} - b_{1}) [∵ 2$ Surfaces $]$
$Δ A = 2 \times 10 (0.6 - 0.5)$
$Δ A = 2 \times 10^{- 4} m^{2}$
We know that,
$∵$ Surface tension $(T) = 7.2 \times 10^{- 2} N / m$
$∴$ Work done $(W) = T \times Δ A$
$W = 7.2 \times 10^{- 2} \times 2 \times 10^{- 4}$
$W = 14.4 \times 10^{- 6}$
$W = 1.44 \times 10^{- 5} N . m$

MHT-CET 2020

Mechanical Properties of Fluids

142829 If the surface tension of a soap solution is $3 \times$ $10^{- 2} N / m$ , then the work done in forming a soap film of $20 cm \times 5 cm$ will be

1

6 \times 10^{- 3} J

2

6 J

3

6 \times 10^{- 4} J

4

6 \times 10^{- 2} J

Explanation:

C Given that,
Total surface area in which soap film is formed $A = 2 \times 20 \times 5$ [Soap bubble form 2 surface Area]
$A = 200 {cm}^{2}$
$A = 200 \times 10^{- 4} m^{2}$
$A = 2 \times 10^{- 2} m^{2}$
Surface tension $(T) = 3 \times 10^{- 2} N / m$
Work done $(W) = T \times A$
$W = 3 \times 10^{- 2} \times 2 \times 10^{- 2}$
$W = 6 \times 10^{- 4} J$

MHT-CET 2020

Mechanical Properties of Fluids

142825 The surface tension of a liquid is $5 N / m$ . If a film is held on a ring of area $0.02 m^{2}$ , its total surface energy is about :

1

2 \times 10^{- 2} J

2

2.5 \times 10^{- 2} J

3

2 \times 10^{- 1} J

4

3 \times 10^{- 1} J

Explanation:

C Given that,
$T = 5 N / m$
$A = 0.02 m^{2}$
We know that,
Work done or surface energy of the film $(E) = 2 T$ A
$= 2 \times 5 \times 0.02$
$= 2 \times 5 \times 0.02$
$= 2 \times 10^{- 1} J$

BCECE-2005

Mechanical Properties of Fluids

142826 Pressure inside two soap bubbles are 1.01 and 1.02 atm. Ratio between their volumes is :

1

102 : 101

2

(102)^{3} : (103)^{3}

3

8 : 1

4

2 : 1

Explanation:

C Given that,
Pressure inside first bubble $= 1.01 atm$
Pressure inside second bubble $= 1.02 atm$
Since atmospheric pressure $= 1 atm$ So,
Excess pressure $Δ P_{1} = 1.01 - 1 = 0.01 atm$
Excess pressure $Δ P_{2} = 1.02 - 1 = 0.02 atm$
Change in pressure,
$Δ P = \frac{4 T}{r}$
$∴ Δ P \propto \frac{1}{r}$
$\frac{Δ P_{2}}{Δ P_{1}} = \frac{r_{1}}{r_{2}}$
$\frac{0.02}{0.01} = \frac{r_{1}}{r_{2}}$
$\frac{r_{1}}{r_{2}} = \frac{2}{1}$
Now,
$V = \frac{4}{3} π r^{3}$
$∴ \frac{V_{1}}{V_{2}} = {(\frac{r_{1}}{r_{2}})}^{3} = {(\frac{2}{1})}^{3} = \frac{8}{1}$
$V_{1} : V_{2} = 8 : 1$

BCECE-2003

Mechanical Properties of Fluids

142827 A water film is formed between the two straight parallel wires, each of length $10 cm$ , kept at a separation of $0.5 cm$ . Now, the separation between them is increased by $1 mm$ without breaking the water film. The work done for this is
( Surface tension of water $= 7.2 \times 10^{- 2} {Nm}^{- 1}$ )

1

1.44 \times 10^{- 5} J

2

5.76 \times 10^{- 5} J

3

7.22 \times 10^{- 6} J

4

2.88 \times 10^{- 5} J

Explanation:

A Given that,
Length of wire, $l = 10 cm$
Separation between them, $b_{1} = 0.5 cm$
If the separation increased by $= 1 mm = 0.1 cm$
$b_{2} = b_{1} + 0.5 = 0.1 + 0.5 = 0.6 cm$
$Δ A = (A_{2} - A_{1})$
$Δ A = 2 l (b_{2} - b_{1}) [∵ 2$ Surfaces $]$
$Δ A = 2 \times 10 (0.6 - 0.5)$
$Δ A = 2 \times 10^{- 4} m^{2}$
We know that,
$∵$ Surface tension $(T) = 7.2 \times 10^{- 2} N / m$
$∴$ Work done $(W) = T \times Δ A$
$W = 7.2 \times 10^{- 2} \times 2 \times 10^{- 4}$
$W = 14.4 \times 10^{- 6}$
$W = 1.44 \times 10^{- 5} N . m$

MHT-CET 2020

Mechanical Properties of Fluids

142829 If the surface tension of a soap solution is $3 \times$ $10^{- 2} N / m$ , then the work done in forming a soap film of $20 cm \times 5 cm$ will be

1

6 \times 10^{- 3} J

2

6 J

3

6 \times 10^{- 4} J

4

6 \times 10^{- 2} J

Explanation:

C Given that,
Total surface area in which soap film is formed $A = 2 \times 20 \times 5$ [Soap bubble form 2 surface Area]
$A = 200 {cm}^{2}$
$A = 200 \times 10^{- 4} m^{2}$
$A = 2 \times 10^{- 2} m^{2}$
Surface tension $(T) = 3 \times 10^{- 2} N / m$
Work done $(W) = T \times A$
$W = 3 \times 10^{- 2} \times 2 \times 10^{- 2}$
$W = 6 \times 10^{- 4} J$

MHT-CET 2020

Mechanical Properties of Fluids

142825 The surface tension of a liquid is $5 N / m$ . If a film is held on a ring of area $0.02 m^{2}$ , its total surface energy is about :

1

2 \times 10^{- 2} J

2

2.5 \times 10^{- 2} J

3

2 \times 10^{- 1} J

4

3 \times 10^{- 1} J

Explanation:

C Given that,
$T = 5 N / m$
$A = 0.02 m^{2}$
We know that,
Work done or surface energy of the film $(E) = 2 T$ A
$= 2 \times 5 \times 0.02$
$= 2 \times 5 \times 0.02$
$= 2 \times 10^{- 1} J$

BCECE-2005

Mechanical Properties of Fluids

142826 Pressure inside two soap bubbles are 1.01 and 1.02 atm. Ratio between their volumes is :

1

102 : 101

2

(102)^{3} : (103)^{3}

3

8 : 1

4

2 : 1

Explanation:

C Given that,
Pressure inside first bubble $= 1.01 atm$
Pressure inside second bubble $= 1.02 atm$
Since atmospheric pressure $= 1 atm$ So,
Excess pressure $Δ P_{1} = 1.01 - 1 = 0.01 atm$
Excess pressure $Δ P_{2} = 1.02 - 1 = 0.02 atm$
Change in pressure,
$Δ P = \frac{4 T}{r}$
$∴ Δ P \propto \frac{1}{r}$
$\frac{Δ P_{2}}{Δ P_{1}} = \frac{r_{1}}{r_{2}}$
$\frac{0.02}{0.01} = \frac{r_{1}}{r_{2}}$
$\frac{r_{1}}{r_{2}} = \frac{2}{1}$
Now,
$V = \frac{4}{3} π r^{3}$
$∴ \frac{V_{1}}{V_{2}} = {(\frac{r_{1}}{r_{2}})}^{3} = {(\frac{2}{1})}^{3} = \frac{8}{1}$
$V_{1} : V_{2} = 8 : 1$

BCECE-2003

Mechanical Properties of Fluids

142827 A water film is formed between the two straight parallel wires, each of length $10 cm$ , kept at a separation of $0.5 cm$ . Now, the separation between them is increased by $1 mm$ without breaking the water film. The work done for this is
( Surface tension of water $= 7.2 \times 10^{- 2} {Nm}^{- 1}$ )

1

1.44 \times 10^{- 5} J

2

5.76 \times 10^{- 5} J

3

7.22 \times 10^{- 6} J

4

2.88 \times 10^{- 5} J

Explanation:

A Given that,
Length of wire, $l = 10 cm$
Separation between them, $b_{1} = 0.5 cm$
If the separation increased by $= 1 mm = 0.1 cm$
$b_{2} = b_{1} + 0.5 = 0.1 + 0.5 = 0.6 cm$
$Δ A = (A_{2} - A_{1})$
$Δ A = 2 l (b_{2} - b_{1}) [∵ 2$ Surfaces $]$
$Δ A = 2 \times 10 (0.6 - 0.5)$
$Δ A = 2 \times 10^{- 4} m^{2}$
We know that,
$∵$ Surface tension $(T) = 7.2 \times 10^{- 2} N / m$
$∴$ Work done $(W) = T \times Δ A$
$W = 7.2 \times 10^{- 2} \times 2 \times 10^{- 4}$
$W = 14.4 \times 10^{- 6}$
$W = 1.44 \times 10^{- 5} N . m$

MHT-CET 2020

Mechanical Properties of Fluids

142829 If the surface tension of a soap solution is $3 \times$ $10^{- 2} N / m$ , then the work done in forming a soap film of $20 cm \times 5 cm$ will be

1

6 \times 10^{- 3} J

2

6 J

3

6 \times 10^{- 4} J

4

6 \times 10^{- 2} J

Explanation:

C Given that,
Total surface area in which soap film is formed $A = 2 \times 20 \times 5$ [Soap bubble form 2 surface Area]
$A = 200 {cm}^{2}$
$A = 200 \times 10^{- 4} m^{2}$
$A = 2 \times 10^{- 2} m^{2}$
Surface tension $(T) = 3 \times 10^{- 2} N / m$
Work done $(W) = T \times A$
$W = 3 \times 10^{- 2} \times 2 \times 10^{- 2}$
$W = 6 \times 10^{- 4} J$

MHT-CET 2020