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Mechanical Properties of Fluids

142830 One thousand small water drops of equal radii combine to form a big drop. The ratio of final surface energy to the total initial surface energy is

1

1 : 1000

2

1 : 10

3

1 : 100

4

1 : 1

Explanation:

B Given,
Let, Radius of small water drop $= r$
Radius of big water drop $= R$
According to question,
Volume of big drop $= 1000 \times$ volume of small drops.
$\frac{4}{3} π R^{3} = \frac{4}{3} π r^{3} \times 1000$
$\frac{R^{3}}{r^{3}} = 1000$
${(\frac{R}{r})}^{3} = (10)^{3}$
$\frac{R}{r} = 10$
We know that,
$Surface energy of drop (U) = A \times T$
For big drop -
$U_{1} = 4 π R^{2} T$
Surface energy of small drops -
$U_{2} = 1000 \times 4 π r^{2} T$
Ratio of final surface energy to the total initial surface energy,
$\frac{U_{1}}{U_{2}} = \frac{4 π T \times R^{2}}{1000 \times 4 π T \times r^{2}}$
$\frac{U_{1}}{U_{2}} = {(\frac{R}{r})}^{2} \times \frac{1}{1000}$
$\frac{U_{1}}{U_{2}} = (10)^{2} \times \frac{1}{1000}$
$\frac{U_{1}}{U_{2}} = \frac{100}{1000}$
$\frac{U_{1}}{U_{2}} = \frac{1}{10}$
$or U_{1} : U_{2} = 1 : 10$

MHT-CET 2020

Mechanical Properties of Fluids

142831 The excess pressure inside a spherical drop of water is three time that of another drop of water. The ratio of their surface area is

1

3 : 1

2

6 : 1

3

1 : 3

4

1 : 9

Explanation:

D Given that,
Pressure inside a spherical drop $(P_{1}) = 3 P_{2}$
Pressure inside drop water $= P_{2}$
We know that,
$P = \frac{2 T}{r}$
According to question $P_{1} = 3 P_{2}$
$\frac{2 T}{r_{1}} = 3 \times \frac{2 T}{r_{2}}$
$r_{2} = 3 r_{1}$
$A = 4 π r^{2}$
$\frac{A_{1}}{A_{2}} = \frac{4 π r_{1}^{2}}{4 π r_{2}^{2}} = {(\frac{r_{1}}{r_{2}})}^{2} (r_{1} = \frac{r_{2}}{3})$
$\frac{A_{1}}{A_{2}} = {(\frac{r_{2}}{3 r_{2}})}^{2}$
$\frac{A_{1}}{A_{2}} = \frac{1}{9}$

MHT-CET 2020

Mechanical Properties of Fluids

142832 Two small drops of mercury each of radius ' $R$ ' coalesce to form a large single drop. The ratio of the total surface energies before and after the change is

1

2^{2 / 3} : 1

2

2^{1 / 3} : 1

3

2 : 1

4

\sqrt{2} : 1

Explanation:

B Given that, radius of large single drops $= R^{'}$ According to question,
Volume of two small drops $=$ volume of single largedrop
$2 V = V^{'}$
$2 \times \frac{4}{3} π R^{3} = \frac{4}{3} π R^{' 3}$
$2 R^{3} = R^{' 3}$
${(\frac{R}{R^{'}})}^{3} = \frac{1}{2} \Rightarrow (\frac{R}{R^{'}}) = {(\frac{1}{2})}^{1 / 3}$
$∴$ We know that surface energy for water drops-
$U = 4 π R^{2} \times T$
According to question-
$\frac{U_{1}}{U_{2}} = \frac{2 \times 4 π R^{2} \times T}{4 π R^{' 2} \times T} = 2 {(\frac{R}{R^{'}})}^{2}$
$\frac{U_{1}}{U_{2}} = 2 \times {[{(\frac{1}{2})}^{1 / 3}]}^{2}$
$\frac{U_{1}}{U_{2}} = 2^{1} \times \frac{1}{2^{2 / 3}} = 2^{1} \times 2^{- 2 / 3}$
$\frac{U_{1}}{U_{2}} = 2^{(1 - \frac{2}{3})} = 2^{(\frac{3 - 2}{3})}$
$\frac{U_{1}}{U_{2}} = 2^{1 / 3}$
But officially option (a) in correct.

MHT-CET 2020

Mechanical Properties of Fluids

142833 Under isothermal conditions, two soap bubbles of radii ' $r_{1}$ ' and ' $r_{2}$ ' coalesce to form a big drop. The radius of the big drop is

1

{(r_{1} - r_{2})}^{1 / 2}

2

{(r_{1} + r_{2})}^{1 / 2}

3

{(r_{1}^{2} + r_{2}^{2})}^{1 / 2}

4

{(r_{1}^{2} - r_{2}^{2})}^{1 / 2}

Explanation:

C Given that,
Radius of first bubbles $= r_{1}$
Radius of second bubbles $= r_{2}$
$∴$ The excess pressure $(P_{1})$ developed inside the first bubble is given by-
$P_{1} = \frac{4 T}{r_{1}} .$
Again, for the $2^{nd}$ bubble-
$P_{2} = \frac{4 T}{r_{2}}$
Let the radius of final bubble is ' $R$ ' then the pressure inside it-
$P = \frac{4 T}{R} \dots \dots$
According to question bubbles are coalesce in isothermal condition-
$∴ PV = P_{1} V_{1} + P_{2} V_{2}$
$(\frac{4 T}{R}) \times \frac{4}{3} π R^{3} = (\frac{4 T}{r_{1}}) \times \frac{4}{3} π r_{1}^{3} + (\frac{4 T}{r_{2}}) \times \frac{4}{3} π r_{2}^{3}$
$(4 T \times \frac{4}{3} π) \times \frac{R^{3}}{R} = 4 T \times \frac{4 π}{3} [\frac{r_{1}^{3}}{r_{1}} + \frac{r_{2}^{3}}{r_{2}}]$
$R^{2} = r_{1}^{2} + r_{2}^{2}$
$R = {(r_{1}^{2} + r_{2}^{2})}^{1 / 2}$

MHT-CET 2020

Mechanical Properties of Fluids

142830 One thousand small water drops of equal radii combine to form a big drop. The ratio of final surface energy to the total initial surface energy is

1

1 : 1000

2

1 : 10

3

1 : 100

4

1 : 1

Explanation:

B Given,
Let, Radius of small water drop $= r$
Radius of big water drop $= R$
According to question,
Volume of big drop $= 1000 \times$ volume of small drops.
$\frac{4}{3} π R^{3} = \frac{4}{3} π r^{3} \times 1000$
$\frac{R^{3}}{r^{3}} = 1000$
${(\frac{R}{r})}^{3} = (10)^{3}$
$\frac{R}{r} = 10$
We know that,
$Surface energy of drop (U) = A \times T$
For big drop -
$U_{1} = 4 π R^{2} T$
Surface energy of small drops -
$U_{2} = 1000 \times 4 π r^{2} T$
Ratio of final surface energy to the total initial surface energy,
$\frac{U_{1}}{U_{2}} = \frac{4 π T \times R^{2}}{1000 \times 4 π T \times r^{2}}$
$\frac{U_{1}}{U_{2}} = {(\frac{R}{r})}^{2} \times \frac{1}{1000}$
$\frac{U_{1}}{U_{2}} = (10)^{2} \times \frac{1}{1000}$
$\frac{U_{1}}{U_{2}} = \frac{100}{1000}$
$\frac{U_{1}}{U_{2}} = \frac{1}{10}$
$or U_{1} : U_{2} = 1 : 10$

MHT-CET 2020

Mechanical Properties of Fluids

142831 The excess pressure inside a spherical drop of water is three time that of another drop of water. The ratio of their surface area is

1

3 : 1

2

6 : 1

3

1 : 3

4

1 : 9

Explanation:

D Given that,
Pressure inside a spherical drop $(P_{1}) = 3 P_{2}$
Pressure inside drop water $= P_{2}$
We know that,
$P = \frac{2 T}{r}$
According to question $P_{1} = 3 P_{2}$
$\frac{2 T}{r_{1}} = 3 \times \frac{2 T}{r_{2}}$
$r_{2} = 3 r_{1}$
$A = 4 π r^{2}$
$\frac{A_{1}}{A_{2}} = \frac{4 π r_{1}^{2}}{4 π r_{2}^{2}} = {(\frac{r_{1}}{r_{2}})}^{2} (r_{1} = \frac{r_{2}}{3})$
$\frac{A_{1}}{A_{2}} = {(\frac{r_{2}}{3 r_{2}})}^{2}$
$\frac{A_{1}}{A_{2}} = \frac{1}{9}$

MHT-CET 2020

Mechanical Properties of Fluids

142832 Two small drops of mercury each of radius ' $R$ ' coalesce to form a large single drop. The ratio of the total surface energies before and after the change is

1

2^{2 / 3} : 1

2

2^{1 / 3} : 1

3

2 : 1

4

\sqrt{2} : 1

Explanation:

B Given that, radius of large single drops $= R^{'}$ According to question,
Volume of two small drops $=$ volume of single largedrop
$2 V = V^{'}$
$2 \times \frac{4}{3} π R^{3} = \frac{4}{3} π R^{' 3}$
$2 R^{3} = R^{' 3}$
${(\frac{R}{R^{'}})}^{3} = \frac{1}{2} \Rightarrow (\frac{R}{R^{'}}) = {(\frac{1}{2})}^{1 / 3}$
$∴$ We know that surface energy for water drops-
$U = 4 π R^{2} \times T$
According to question-
$\frac{U_{1}}{U_{2}} = \frac{2 \times 4 π R^{2} \times T}{4 π R^{' 2} \times T} = 2 {(\frac{R}{R^{'}})}^{2}$
$\frac{U_{1}}{U_{2}} = 2 \times {[{(\frac{1}{2})}^{1 / 3}]}^{2}$
$\frac{U_{1}}{U_{2}} = 2^{1} \times \frac{1}{2^{2 / 3}} = 2^{1} \times 2^{- 2 / 3}$
$\frac{U_{1}}{U_{2}} = 2^{(1 - \frac{2}{3})} = 2^{(\frac{3 - 2}{3})}$
$\frac{U_{1}}{U_{2}} = 2^{1 / 3}$
But officially option (a) in correct.

MHT-CET 2020

Mechanical Properties of Fluids

142833 Under isothermal conditions, two soap bubbles of radii ' $r_{1}$ ' and ' $r_{2}$ ' coalesce to form a big drop. The radius of the big drop is

1

{(r_{1} - r_{2})}^{1 / 2}

2

{(r_{1} + r_{2})}^{1 / 2}

3

{(r_{1}^{2} + r_{2}^{2})}^{1 / 2}

4

{(r_{1}^{2} - r_{2}^{2})}^{1 / 2}

Explanation:

C Given that,
Radius of first bubbles $= r_{1}$
Radius of second bubbles $= r_{2}$
$∴$ The excess pressure $(P_{1})$ developed inside the first bubble is given by-
$P_{1} = \frac{4 T}{r_{1}} .$
Again, for the $2^{nd}$ bubble-
$P_{2} = \frac{4 T}{r_{2}}$
Let the radius of final bubble is ' $R$ ' then the pressure inside it-
$P = \frac{4 T}{R} \dots \dots$
According to question bubbles are coalesce in isothermal condition-
$∴ PV = P_{1} V_{1} + P_{2} V_{2}$
$(\frac{4 T}{R}) \times \frac{4}{3} π R^{3} = (\frac{4 T}{r_{1}}) \times \frac{4}{3} π r_{1}^{3} + (\frac{4 T}{r_{2}}) \times \frac{4}{3} π r_{2}^{3}$
$(4 T \times \frac{4}{3} π) \times \frac{R^{3}}{R} = 4 T \times \frac{4 π}{3} [\frac{r_{1}^{3}}{r_{1}} + \frac{r_{2}^{3}}{r_{2}}]$
$R^{2} = r_{1}^{2} + r_{2}^{2}$
$R = {(r_{1}^{2} + r_{2}^{2})}^{1 / 2}$

MHT-CET 2020

Mechanical Properties of Fluids

142830 One thousand small water drops of equal radii combine to form a big drop. The ratio of final surface energy to the total initial surface energy is

1

1 : 1000

2

1 : 10

3

1 : 100

4

1 : 1

Explanation:

B Given,
Let, Radius of small water drop $= r$
Radius of big water drop $= R$
According to question,
Volume of big drop $= 1000 \times$ volume of small drops.
$\frac{4}{3} π R^{3} = \frac{4}{3} π r^{3} \times 1000$
$\frac{R^{3}}{r^{3}} = 1000$
${(\frac{R}{r})}^{3} = (10)^{3}$
$\frac{R}{r} = 10$
We know that,
$Surface energy of drop (U) = A \times T$
For big drop -
$U_{1} = 4 π R^{2} T$
Surface energy of small drops -
$U_{2} = 1000 \times 4 π r^{2} T$
Ratio of final surface energy to the total initial surface energy,
$\frac{U_{1}}{U_{2}} = \frac{4 π T \times R^{2}}{1000 \times 4 π T \times r^{2}}$
$\frac{U_{1}}{U_{2}} = {(\frac{R}{r})}^{2} \times \frac{1}{1000}$
$\frac{U_{1}}{U_{2}} = (10)^{2} \times \frac{1}{1000}$
$\frac{U_{1}}{U_{2}} = \frac{100}{1000}$
$\frac{U_{1}}{U_{2}} = \frac{1}{10}$
$or U_{1} : U_{2} = 1 : 10$

MHT-CET 2020

Mechanical Properties of Fluids

142831 The excess pressure inside a spherical drop of water is three time that of another drop of water. The ratio of their surface area is

1

3 : 1

2

6 : 1

3

1 : 3

4

1 : 9

Explanation:

D Given that,
Pressure inside a spherical drop $(P_{1}) = 3 P_{2}$
Pressure inside drop water $= P_{2}$
We know that,
$P = \frac{2 T}{r}$
According to question $P_{1} = 3 P_{2}$
$\frac{2 T}{r_{1}} = 3 \times \frac{2 T}{r_{2}}$
$r_{2} = 3 r_{1}$
$A = 4 π r^{2}$
$\frac{A_{1}}{A_{2}} = \frac{4 π r_{1}^{2}}{4 π r_{2}^{2}} = {(\frac{r_{1}}{r_{2}})}^{2} (r_{1} = \frac{r_{2}}{3})$
$\frac{A_{1}}{A_{2}} = {(\frac{r_{2}}{3 r_{2}})}^{2}$
$\frac{A_{1}}{A_{2}} = \frac{1}{9}$

MHT-CET 2020

Mechanical Properties of Fluids

142832 Two small drops of mercury each of radius ' $R$ ' coalesce to form a large single drop. The ratio of the total surface energies before and after the change is

1

2^{2 / 3} : 1

2

2^{1 / 3} : 1

3

2 : 1

4

\sqrt{2} : 1

Explanation:

B Given that, radius of large single drops $= R^{'}$ According to question,
Volume of two small drops $=$ volume of single largedrop
$2 V = V^{'}$
$2 \times \frac{4}{3} π R^{3} = \frac{4}{3} π R^{' 3}$
$2 R^{3} = R^{' 3}$
${(\frac{R}{R^{'}})}^{3} = \frac{1}{2} \Rightarrow (\frac{R}{R^{'}}) = {(\frac{1}{2})}^{1 / 3}$
$∴$ We know that surface energy for water drops-
$U = 4 π R^{2} \times T$
According to question-
$\frac{U_{1}}{U_{2}} = \frac{2 \times 4 π R^{2} \times T}{4 π R^{' 2} \times T} = 2 {(\frac{R}{R^{'}})}^{2}$
$\frac{U_{1}}{U_{2}} = 2 \times {[{(\frac{1}{2})}^{1 / 3}]}^{2}$
$\frac{U_{1}}{U_{2}} = 2^{1} \times \frac{1}{2^{2 / 3}} = 2^{1} \times 2^{- 2 / 3}$
$\frac{U_{1}}{U_{2}} = 2^{(1 - \frac{2}{3})} = 2^{(\frac{3 - 2}{3})}$
$\frac{U_{1}}{U_{2}} = 2^{1 / 3}$
But officially option (a) in correct.

MHT-CET 2020

Mechanical Properties of Fluids

142833 Under isothermal conditions, two soap bubbles of radii ' $r_{1}$ ' and ' $r_{2}$ ' coalesce to form a big drop. The radius of the big drop is

1

{(r_{1} - r_{2})}^{1 / 2}

2

{(r_{1} + r_{2})}^{1 / 2}

3

{(r_{1}^{2} + r_{2}^{2})}^{1 / 2}

4

{(r_{1}^{2} - r_{2}^{2})}^{1 / 2}

Explanation:

C Given that,
Radius of first bubbles $= r_{1}$
Radius of second bubbles $= r_{2}$
$∴$ The excess pressure $(P_{1})$ developed inside the first bubble is given by-
$P_{1} = \frac{4 T}{r_{1}} .$
Again, for the $2^{nd}$ bubble-
$P_{2} = \frac{4 T}{r_{2}}$
Let the radius of final bubble is ' $R$ ' then the pressure inside it-
$P = \frac{4 T}{R} \dots \dots$
According to question bubbles are coalesce in isothermal condition-
$∴ PV = P_{1} V_{1} + P_{2} V_{2}$
$(\frac{4 T}{R}) \times \frac{4}{3} π R^{3} = (\frac{4 T}{r_{1}}) \times \frac{4}{3} π r_{1}^{3} + (\frac{4 T}{r_{2}}) \times \frac{4}{3} π r_{2}^{3}$
$(4 T \times \frac{4}{3} π) \times \frac{R^{3}}{R} = 4 T \times \frac{4 π}{3} [\frac{r_{1}^{3}}{r_{1}} + \frac{r_{2}^{3}}{r_{2}}]$
$R^{2} = r_{1}^{2} + r_{2}^{2}$
$R = {(r_{1}^{2} + r_{2}^{2})}^{1 / 2}$

MHT-CET 2020

Mechanical Properties of Fluids

142830 One thousand small water drops of equal radii combine to form a big drop. The ratio of final surface energy to the total initial surface energy is

1

1 : 1000

2

1 : 10

3

1 : 100

4

1 : 1

Explanation:

B Given,
Let, Radius of small water drop $= r$
Radius of big water drop $= R$
According to question,
Volume of big drop $= 1000 \times$ volume of small drops.
$\frac{4}{3} π R^{3} = \frac{4}{3} π r^{3} \times 1000$
$\frac{R^{3}}{r^{3}} = 1000$
${(\frac{R}{r})}^{3} = (10)^{3}$
$\frac{R}{r} = 10$
We know that,
$Surface energy of drop (U) = A \times T$
For big drop -
$U_{1} = 4 π R^{2} T$
Surface energy of small drops -
$U_{2} = 1000 \times 4 π r^{2} T$
Ratio of final surface energy to the total initial surface energy,
$\frac{U_{1}}{U_{2}} = \frac{4 π T \times R^{2}}{1000 \times 4 π T \times r^{2}}$
$\frac{U_{1}}{U_{2}} = {(\frac{R}{r})}^{2} \times \frac{1}{1000}$
$\frac{U_{1}}{U_{2}} = (10)^{2} \times \frac{1}{1000}$
$\frac{U_{1}}{U_{2}} = \frac{100}{1000}$
$\frac{U_{1}}{U_{2}} = \frac{1}{10}$
$or U_{1} : U_{2} = 1 : 10$

MHT-CET 2020

Mechanical Properties of Fluids

142831 The excess pressure inside a spherical drop of water is three time that of another drop of water. The ratio of their surface area is

1

3 : 1

2

6 : 1

3

1 : 3

4

1 : 9

Explanation:

D Given that,
Pressure inside a spherical drop $(P_{1}) = 3 P_{2}$
Pressure inside drop water $= P_{2}$
We know that,
$P = \frac{2 T}{r}$
According to question $P_{1} = 3 P_{2}$
$\frac{2 T}{r_{1}} = 3 \times \frac{2 T}{r_{2}}$
$r_{2} = 3 r_{1}$
$A = 4 π r^{2}$
$\frac{A_{1}}{A_{2}} = \frac{4 π r_{1}^{2}}{4 π r_{2}^{2}} = {(\frac{r_{1}}{r_{2}})}^{2} (r_{1} = \frac{r_{2}}{3})$
$\frac{A_{1}}{A_{2}} = {(\frac{r_{2}}{3 r_{2}})}^{2}$
$\frac{A_{1}}{A_{2}} = \frac{1}{9}$

MHT-CET 2020

Mechanical Properties of Fluids

142832 Two small drops of mercury each of radius ' $R$ ' coalesce to form a large single drop. The ratio of the total surface energies before and after the change is

1

2^{2 / 3} : 1

2

2^{1 / 3} : 1

3

2 : 1

4

\sqrt{2} : 1

Explanation:

B Given that, radius of large single drops $= R^{'}$ According to question,
Volume of two small drops $=$ volume of single largedrop
$2 V = V^{'}$
$2 \times \frac{4}{3} π R^{3} = \frac{4}{3} π R^{' 3}$
$2 R^{3} = R^{' 3}$
${(\frac{R}{R^{'}})}^{3} = \frac{1}{2} \Rightarrow (\frac{R}{R^{'}}) = {(\frac{1}{2})}^{1 / 3}$
$∴$ We know that surface energy for water drops-
$U = 4 π R^{2} \times T$
According to question-
$\frac{U_{1}}{U_{2}} = \frac{2 \times 4 π R^{2} \times T}{4 π R^{' 2} \times T} = 2 {(\frac{R}{R^{'}})}^{2}$
$\frac{U_{1}}{U_{2}} = 2 \times {[{(\frac{1}{2})}^{1 / 3}]}^{2}$
$\frac{U_{1}}{U_{2}} = 2^{1} \times \frac{1}{2^{2 / 3}} = 2^{1} \times 2^{- 2 / 3}$
$\frac{U_{1}}{U_{2}} = 2^{(1 - \frac{2}{3})} = 2^{(\frac{3 - 2}{3})}$
$\frac{U_{1}}{U_{2}} = 2^{1 / 3}$
But officially option (a) in correct.

MHT-CET 2020

Mechanical Properties of Fluids

142833 Under isothermal conditions, two soap bubbles of radii ' $r_{1}$ ' and ' $r_{2}$ ' coalesce to form a big drop. The radius of the big drop is

1

{(r_{1} - r_{2})}^{1 / 2}

2

{(r_{1} + r_{2})}^{1 / 2}

3

{(r_{1}^{2} + r_{2}^{2})}^{1 / 2}

4

{(r_{1}^{2} - r_{2}^{2})}^{1 / 2}

Explanation:

C Given that,
Radius of first bubbles $= r_{1}$
Radius of second bubbles $= r_{2}$
$∴$ The excess pressure $(P_{1})$ developed inside the first bubble is given by-
$P_{1} = \frac{4 T}{r_{1}} .$
Again, for the $2^{nd}$ bubble-
$P_{2} = \frac{4 T}{r_{2}}$
Let the radius of final bubble is ' $R$ ' then the pressure inside it-
$P = \frac{4 T}{R} \dots \dots$
According to question bubbles are coalesce in isothermal condition-
$∴ PV = P_{1} V_{1} + P_{2} V_{2}$
$(\frac{4 T}{R}) \times \frac{4}{3} π R^{3} = (\frac{4 T}{r_{1}}) \times \frac{4}{3} π r_{1}^{3} + (\frac{4 T}{r_{2}}) \times \frac{4}{3} π r_{2}^{3}$
$(4 T \times \frac{4}{3} π) \times \frac{R^{3}}{R} = 4 T \times \frac{4 π}{3} [\frac{r_{1}^{3}}{r_{1}} + \frac{r_{2}^{3}}{r_{2}}]$
$R^{2} = r_{1}^{2} + r_{2}^{2}$
$R = {(r_{1}^{2} + r_{2}^{2})}^{1 / 2}$

MHT-CET 2020

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