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Mechanical Properties of Fluids

142834 When a mercury drop of a radius ' $R$ ', breaks into ' $n$ ' droplets of equal size, the radius ' $r$ ' of each droplets is

1

r = \frac{R}{\sqrt{n}}

2

r = {Rn}^{\frac{1}{3}}

3

r = \frac{R}{n}

4

r = \frac{R}{n^{1 / 3}}

Explanation:

D The volume of $Hg$ drops before breaks and after breaks will be same.
$∴ \frac{4}{3} π R^{3} = n \times \frac{4}{3} π r^{3}$
$R^{3} = {nr}^{3}$
$\Rightarrow {(\frac{R}{r})}^{3} = n$
$\Rightarrow \frac{R}{r} = n^{1 / 3}$
$r = \frac{R}{n^{1 / 3}}$

MHT-CET 2020

Mechanical Properties of Fluids

142836 The work done in blowing a soap bubble of radius ' $R$ ' is ' $W_{1}$ ' at room temperature. Now the soap solution is heated. From the heated solution another soap bubble of radius ' $2 R$ ' is blown and the work done is ' $W_{2}$ ', Then

1

W_{2} = 0

2

W_{2} < 4 W_{1}

3

W_{2} = 4 W_{1}

4

W_{2} = W_{1}

Explanation:

B The work done to blowing a soap bubble of radius ' $R$ ' and surface tension ' $T$ ' is given as-
$W_{1} = 2 \times 4 π R^{2} \times T = 8 π R^{2} T$
Now, if radius is doubled then work done to blowing the soap bubble is-
$W_{2} = 2 \times 4 π (2 R)^{2} \times T^{'}$
$W_{2} = 4 \times 8 π R^{2} \times 4 T^{'}$
Dividing equation (ii) by (i), we get-
$\frac{W_{2}}{W_{1}} = \frac{4 T^{'}}{T}$
But, in the question soap bubbles is heated then it's surface tension is decrease so, $W_{2} < 4 W_{1}$

MHT-CET 2020

Mechanical Properties of Fluids

142837 A fix number of spherical drops of a liquid of radius ' $r$ ' coalesce to form a large drop of radius ' $R$ ' and volume ' $V$ '. If ' $T$ ' is the surface tension then energy

1

4 VT (\frac{1}{r} - \frac{1}{R})

is released

2

3 VT (\frac{1}{r} - \frac{1}{R})

is released

3 is neither released nor absorbed

4

3 VT (\frac{1}{r} - \frac{1}{R})

is absorbed

Explanation:

B Let the number of small spherical drops be ' $n$ ' and radius 'r'.
After coalesce it forms a big drop having radius ' $R$ '
$∵$ Volume of drops before coalesce will be same after coalesce.
$n \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3}$
${nr}^{3} = R^{3}$
$n = \frac{R^{3}}{r^{3}}$
When the number of drops coalesce the energy is released.
$∴$ Initial energy of drops $(E_{1}) = n \times 4 π r^{2} T$
Final energy of drop $(E_{2}) = 4 π R^{2} T$
$∴$ Energy released $(U) = E_{2} - E_{1}$
$U = (4 π R^{2} \cdot T - n \cdot 4 π r^{2} \cdot T)$
$U = 4 π T (R^{2} - n r^{2})$
Putting the value of ' $n$ ', we get -
$U = 4 π T (R^{2} - \frac{R^{3}}{r^{3}} \cdot r^{2})$
$U = 4 π {TR}^{3} [\frac{1}{R} - \frac{1}{r}]$
$U = 3 \times \frac{4}{3} π R^{3} T (\frac{1}{R} - \frac{1}{r})$
$U = 3 VT [\frac{1}{R} - \frac{1}{r}] (∵ V = \frac{4}{3} π R^{3})$
$U = - 3 VT [\frac{1}{r} - \frac{1}{R}]$
Hence, $U = 3 VT [\frac{1}{r} - \frac{1}{R}]$ is released.

MHT-CET 2020

Mechanical Properties of Fluids

142838 A mercury drop of radius ' $R$ ' is divided into 27 droplets of same size. The radius ' $r$ ' of each droplet is

1

r = \frac{R}{3}

2

r = \frac{R}{9}

3

r = \frac{R}{27}

4

r = 3 R

Explanation:

A The volume of mercury drop before dividing into 27 drops will be same to volume of 27 drops.
$∴ \frac{4}{3} π R^{3} = 27 \times \frac{4}{3} π r^{3}$
$R^{3} = 27 r^{3}$
$R^{3} = (3)^{3} r^{3}$
$R = 3 r$
$r = \frac{R}{3}$

MHT-CET 2020

Mechanical Properties of Fluids

142839 Under isothermal conditions, two soap bubbles of radii ' $r_{1}$ ' and ' $r_{2}$ ' combine to form a single soap bubble of radius ' $R$ '. The surface tension of soap solution is $(P =$ outside pressure $)$

1

P (R^{3} - r_{1}^{3} - r_{2}^{3}) / 4 (r_{1}^{2} + r_{2}^{2} - R^{2})

2

P (r_{1}^{3} - R^{3} + r_{2}^{3}) / 4 (R^{2} - r_{1}^{2} - r_{2}^{2})

3

P (R^{3} + r_{1}^{3} + r_{2}^{3}) / 4 (r_{1}^{2} + r_{2}^{2} - R^{2})

4

P (r_{1}^{3} - r_{2}^{3} + R^{3}) / 4 (r_{1}^{2} + r_{2}^{2} + R^{2})

Explanation:

A Pressure inside the first bubble $= P + \frac{4 T}{r_{1}}$
Pressure inside the second bubble $= P + \frac{4 T}{r_{2}}$
We know that, $P V = n R θ$
Where, $θ =$ absolute temperature
As mass of the air inside the soap bubble is conserved $n_{1} = n_{2} = n$
and under isothermal condition, temperature
$θ_{1} = θ_{2} = θ$
$∵ (P + \frac{4 T}{r_{1}}) \frac{4 π}{3} r_{1}^{3} = nR θ$
$And (P + \frac{4 T}{r_{2}}) \frac{4 π}{3} r_{2}^{3} = nR θ$
$∵ (P + \frac{4 T}{R}) \frac{4 π}{3} R^{3} = (P + \frac{4 T}{r_{1}}) \frac{4 π}{3} r_{1}^{3} + (P + \frac{4 T}{r_{2}}) \frac{4 π}{3} r_{2}^{3}$
$(P + \frac{4 T}{R}) R^{3} = (P + \frac{4 T}{r_{1}}) r_{1}^{3} + (P + \frac{4 T}{r_{2}}) r_{2}^{3}$
On solving the equation, we get-
$T = \frac{P (R^{3} - r_{1}^{3} - r_{2}^{3})}{4 (r_{1}^{2} + r_{2}^{2} - R^{2})}$

MHT-CET 2020

Mechanical Properties of Fluids

142834 When a mercury drop of a radius ' $R$ ', breaks into ' $n$ ' droplets of equal size, the radius ' $r$ ' of each droplets is

1

r = \frac{R}{\sqrt{n}}

2

r = {Rn}^{\frac{1}{3}}

3

r = \frac{R}{n}

4

r = \frac{R}{n^{1 / 3}}

Explanation:

D The volume of $Hg$ drops before breaks and after breaks will be same.
$∴ \frac{4}{3} π R^{3} = n \times \frac{4}{3} π r^{3}$
$R^{3} = {nr}^{3}$
$\Rightarrow {(\frac{R}{r})}^{3} = n$
$\Rightarrow \frac{R}{r} = n^{1 / 3}$
$r = \frac{R}{n^{1 / 3}}$

MHT-CET 2020

Mechanical Properties of Fluids

142836 The work done in blowing a soap bubble of radius ' $R$ ' is ' $W_{1}$ ' at room temperature. Now the soap solution is heated. From the heated solution another soap bubble of radius ' $2 R$ ' is blown and the work done is ' $W_{2}$ ', Then

1

W_{2} = 0

2

W_{2} < 4 W_{1}

3

W_{2} = 4 W_{1}

4

W_{2} = W_{1}

Explanation:

B The work done to blowing a soap bubble of radius ' $R$ ' and surface tension ' $T$ ' is given as-
$W_{1} = 2 \times 4 π R^{2} \times T = 8 π R^{2} T$
Now, if radius is doubled then work done to blowing the soap bubble is-
$W_{2} = 2 \times 4 π (2 R)^{2} \times T^{'}$
$W_{2} = 4 \times 8 π R^{2} \times 4 T^{'}$
Dividing equation (ii) by (i), we get-
$\frac{W_{2}}{W_{1}} = \frac{4 T^{'}}{T}$
But, in the question soap bubbles is heated then it's surface tension is decrease so, $W_{2} < 4 W_{1}$

MHT-CET 2020

Mechanical Properties of Fluids

142837 A fix number of spherical drops of a liquid of radius ' $r$ ' coalesce to form a large drop of radius ' $R$ ' and volume ' $V$ '. If ' $T$ ' is the surface tension then energy

1

4 VT (\frac{1}{r} - \frac{1}{R})

is released

2

3 VT (\frac{1}{r} - \frac{1}{R})

is released

3 is neither released nor absorbed

4

3 VT (\frac{1}{r} - \frac{1}{R})

is absorbed

Explanation:

B Let the number of small spherical drops be ' $n$ ' and radius 'r'.
After coalesce it forms a big drop having radius ' $R$ '
$∵$ Volume of drops before coalesce will be same after coalesce.
$n \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3}$
${nr}^{3} = R^{3}$
$n = \frac{R^{3}}{r^{3}}$
When the number of drops coalesce the energy is released.
$∴$ Initial energy of drops $(E_{1}) = n \times 4 π r^{2} T$
Final energy of drop $(E_{2}) = 4 π R^{2} T$
$∴$ Energy released $(U) = E_{2} - E_{1}$
$U = (4 π R^{2} \cdot T - n \cdot 4 π r^{2} \cdot T)$
$U = 4 π T (R^{2} - n r^{2})$
Putting the value of ' $n$ ', we get -
$U = 4 π T (R^{2} - \frac{R^{3}}{r^{3}} \cdot r^{2})$
$U = 4 π {TR}^{3} [\frac{1}{R} - \frac{1}{r}]$
$U = 3 \times \frac{4}{3} π R^{3} T (\frac{1}{R} - \frac{1}{r})$
$U = 3 VT [\frac{1}{R} - \frac{1}{r}] (∵ V = \frac{4}{3} π R^{3})$
$U = - 3 VT [\frac{1}{r} - \frac{1}{R}]$
Hence, $U = 3 VT [\frac{1}{r} - \frac{1}{R}]$ is released.

MHT-CET 2020

Mechanical Properties of Fluids

142838 A mercury drop of radius ' $R$ ' is divided into 27 droplets of same size. The radius ' $r$ ' of each droplet is

1

r = \frac{R}{3}

2

r = \frac{R}{9}

3

r = \frac{R}{27}

4

r = 3 R

Explanation:

A The volume of mercury drop before dividing into 27 drops will be same to volume of 27 drops.
$∴ \frac{4}{3} π R^{3} = 27 \times \frac{4}{3} π r^{3}$
$R^{3} = 27 r^{3}$
$R^{3} = (3)^{3} r^{3}$
$R = 3 r$
$r = \frac{R}{3}$

MHT-CET 2020

Mechanical Properties of Fluids

142839 Under isothermal conditions, two soap bubbles of radii ' $r_{1}$ ' and ' $r_{2}$ ' combine to form a single soap bubble of radius ' $R$ '. The surface tension of soap solution is $(P =$ outside pressure $)$

1

P (R^{3} - r_{1}^{3} - r_{2}^{3}) / 4 (r_{1}^{2} + r_{2}^{2} - R^{2})

2

P (r_{1}^{3} - R^{3} + r_{2}^{3}) / 4 (R^{2} - r_{1}^{2} - r_{2}^{2})

3

P (R^{3} + r_{1}^{3} + r_{2}^{3}) / 4 (r_{1}^{2} + r_{2}^{2} - R^{2})

4

P (r_{1}^{3} - r_{2}^{3} + R^{3}) / 4 (r_{1}^{2} + r_{2}^{2} + R^{2})

Explanation:

A Pressure inside the first bubble $= P + \frac{4 T}{r_{1}}$
Pressure inside the second bubble $= P + \frac{4 T}{r_{2}}$
We know that, $P V = n R θ$
Where, $θ =$ absolute temperature
As mass of the air inside the soap bubble is conserved $n_{1} = n_{2} = n$
and under isothermal condition, temperature
$θ_{1} = θ_{2} = θ$
$∵ (P + \frac{4 T}{r_{1}}) \frac{4 π}{3} r_{1}^{3} = nR θ$
$And (P + \frac{4 T}{r_{2}}) \frac{4 π}{3} r_{2}^{3} = nR θ$
$∵ (P + \frac{4 T}{R}) \frac{4 π}{3} R^{3} = (P + \frac{4 T}{r_{1}}) \frac{4 π}{3} r_{1}^{3} + (P + \frac{4 T}{r_{2}}) \frac{4 π}{3} r_{2}^{3}$
$(P + \frac{4 T}{R}) R^{3} = (P + \frac{4 T}{r_{1}}) r_{1}^{3} + (P + \frac{4 T}{r_{2}}) r_{2}^{3}$
On solving the equation, we get-
$T = \frac{P (R^{3} - r_{1}^{3} - r_{2}^{3})}{4 (r_{1}^{2} + r_{2}^{2} - R^{2})}$

MHT-CET 2020

Mechanical Properties of Fluids

142834 When a mercury drop of a radius ' $R$ ', breaks into ' $n$ ' droplets of equal size, the radius ' $r$ ' of each droplets is

1

r = \frac{R}{\sqrt{n}}

2

r = {Rn}^{\frac{1}{3}}

3

r = \frac{R}{n}

4

r = \frac{R}{n^{1 / 3}}

Explanation:

D The volume of $Hg$ drops before breaks and after breaks will be same.
$∴ \frac{4}{3} π R^{3} = n \times \frac{4}{3} π r^{3}$
$R^{3} = {nr}^{3}$
$\Rightarrow {(\frac{R}{r})}^{3} = n$
$\Rightarrow \frac{R}{r} = n^{1 / 3}$
$r = \frac{R}{n^{1 / 3}}$

MHT-CET 2020

Mechanical Properties of Fluids

142836 The work done in blowing a soap bubble of radius ' $R$ ' is ' $W_{1}$ ' at room temperature. Now the soap solution is heated. From the heated solution another soap bubble of radius ' $2 R$ ' is blown and the work done is ' $W_{2}$ ', Then

1

W_{2} = 0

2

W_{2} < 4 W_{1}

3

W_{2} = 4 W_{1}

4

W_{2} = W_{1}

Explanation:

B The work done to blowing a soap bubble of radius ' $R$ ' and surface tension ' $T$ ' is given as-
$W_{1} = 2 \times 4 π R^{2} \times T = 8 π R^{2} T$
Now, if radius is doubled then work done to blowing the soap bubble is-
$W_{2} = 2 \times 4 π (2 R)^{2} \times T^{'}$
$W_{2} = 4 \times 8 π R^{2} \times 4 T^{'}$
Dividing equation (ii) by (i), we get-
$\frac{W_{2}}{W_{1}} = \frac{4 T^{'}}{T}$
But, in the question soap bubbles is heated then it's surface tension is decrease so, $W_{2} < 4 W_{1}$

MHT-CET 2020

Mechanical Properties of Fluids

142837 A fix number of spherical drops of a liquid of radius ' $r$ ' coalesce to form a large drop of radius ' $R$ ' and volume ' $V$ '. If ' $T$ ' is the surface tension then energy

1

4 VT (\frac{1}{r} - \frac{1}{R})

is released

2

3 VT (\frac{1}{r} - \frac{1}{R})

is released

3 is neither released nor absorbed

4

3 VT (\frac{1}{r} - \frac{1}{R})

is absorbed

Explanation:

B Let the number of small spherical drops be ' $n$ ' and radius 'r'.
After coalesce it forms a big drop having radius ' $R$ '
$∵$ Volume of drops before coalesce will be same after coalesce.
$n \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3}$
${nr}^{3} = R^{3}$
$n = \frac{R^{3}}{r^{3}}$
When the number of drops coalesce the energy is released.
$∴$ Initial energy of drops $(E_{1}) = n \times 4 π r^{2} T$
Final energy of drop $(E_{2}) = 4 π R^{2} T$
$∴$ Energy released $(U) = E_{2} - E_{1}$
$U = (4 π R^{2} \cdot T - n \cdot 4 π r^{2} \cdot T)$
$U = 4 π T (R^{2} - n r^{2})$
Putting the value of ' $n$ ', we get -
$U = 4 π T (R^{2} - \frac{R^{3}}{r^{3}} \cdot r^{2})$
$U = 4 π {TR}^{3} [\frac{1}{R} - \frac{1}{r}]$
$U = 3 \times \frac{4}{3} π R^{3} T (\frac{1}{R} - \frac{1}{r})$
$U = 3 VT [\frac{1}{R} - \frac{1}{r}] (∵ V = \frac{4}{3} π R^{3})$
$U = - 3 VT [\frac{1}{r} - \frac{1}{R}]$
Hence, $U = 3 VT [\frac{1}{r} - \frac{1}{R}]$ is released.

MHT-CET 2020

Mechanical Properties of Fluids

142838 A mercury drop of radius ' $R$ ' is divided into 27 droplets of same size. The radius ' $r$ ' of each droplet is

1

r = \frac{R}{3}

2

r = \frac{R}{9}

3

r = \frac{R}{27}

4

r = 3 R

Explanation:

A The volume of mercury drop before dividing into 27 drops will be same to volume of 27 drops.
$∴ \frac{4}{3} π R^{3} = 27 \times \frac{4}{3} π r^{3}$
$R^{3} = 27 r^{3}$
$R^{3} = (3)^{3} r^{3}$
$R = 3 r$
$r = \frac{R}{3}$

MHT-CET 2020

Mechanical Properties of Fluids

142839 Under isothermal conditions, two soap bubbles of radii ' $r_{1}$ ' and ' $r_{2}$ ' combine to form a single soap bubble of radius ' $R$ '. The surface tension of soap solution is $(P =$ outside pressure $)$

1

P (R^{3} - r_{1}^{3} - r_{2}^{3}) / 4 (r_{1}^{2} + r_{2}^{2} - R^{2})

2

P (r_{1}^{3} - R^{3} + r_{2}^{3}) / 4 (R^{2} - r_{1}^{2} - r_{2}^{2})

3

P (R^{3} + r_{1}^{3} + r_{2}^{3}) / 4 (r_{1}^{2} + r_{2}^{2} - R^{2})

4

P (r_{1}^{3} - r_{2}^{3} + R^{3}) / 4 (r_{1}^{2} + r_{2}^{2} + R^{2})

Explanation:

A Pressure inside the first bubble $= P + \frac{4 T}{r_{1}}$
Pressure inside the second bubble $= P + \frac{4 T}{r_{2}}$
We know that, $P V = n R θ$
Where, $θ =$ absolute temperature
As mass of the air inside the soap bubble is conserved $n_{1} = n_{2} = n$
and under isothermal condition, temperature
$θ_{1} = θ_{2} = θ$
$∵ (P + \frac{4 T}{r_{1}}) \frac{4 π}{3} r_{1}^{3} = nR θ$
$And (P + \frac{4 T}{r_{2}}) \frac{4 π}{3} r_{2}^{3} = nR θ$
$∵ (P + \frac{4 T}{R}) \frac{4 π}{3} R^{3} = (P + \frac{4 T}{r_{1}}) \frac{4 π}{3} r_{1}^{3} + (P + \frac{4 T}{r_{2}}) \frac{4 π}{3} r_{2}^{3}$
$(P + \frac{4 T}{R}) R^{3} = (P + \frac{4 T}{r_{1}}) r_{1}^{3} + (P + \frac{4 T}{r_{2}}) r_{2}^{3}$
On solving the equation, we get-
$T = \frac{P (R^{3} - r_{1}^{3} - r_{2}^{3})}{4 (r_{1}^{2} + r_{2}^{2} - R^{2})}$

MHT-CET 2020

Mechanical Properties of Fluids

142834 When a mercury drop of a radius ' $R$ ', breaks into ' $n$ ' droplets of equal size, the radius ' $r$ ' of each droplets is

1

r = \frac{R}{\sqrt{n}}

2

r = {Rn}^{\frac{1}{3}}

3

r = \frac{R}{n}

4

r = \frac{R}{n^{1 / 3}}

Explanation:

D The volume of $Hg$ drops before breaks and after breaks will be same.
$∴ \frac{4}{3} π R^{3} = n \times \frac{4}{3} π r^{3}$
$R^{3} = {nr}^{3}$
$\Rightarrow {(\frac{R}{r})}^{3} = n$
$\Rightarrow \frac{R}{r} = n^{1 / 3}$
$r = \frac{R}{n^{1 / 3}}$

MHT-CET 2020

Mechanical Properties of Fluids

142836 The work done in blowing a soap bubble of radius ' $R$ ' is ' $W_{1}$ ' at room temperature. Now the soap solution is heated. From the heated solution another soap bubble of radius ' $2 R$ ' is blown and the work done is ' $W_{2}$ ', Then

1

W_{2} = 0

2

W_{2} < 4 W_{1}

3

W_{2} = 4 W_{1}

4

W_{2} = W_{1}

Explanation:

B The work done to blowing a soap bubble of radius ' $R$ ' and surface tension ' $T$ ' is given as-
$W_{1} = 2 \times 4 π R^{2} \times T = 8 π R^{2} T$
Now, if radius is doubled then work done to blowing the soap bubble is-
$W_{2} = 2 \times 4 π (2 R)^{2} \times T^{'}$
$W_{2} = 4 \times 8 π R^{2} \times 4 T^{'}$
Dividing equation (ii) by (i), we get-
$\frac{W_{2}}{W_{1}} = \frac{4 T^{'}}{T}$
But, in the question soap bubbles is heated then it's surface tension is decrease so, $W_{2} < 4 W_{1}$

MHT-CET 2020

Mechanical Properties of Fluids

142837 A fix number of spherical drops of a liquid of radius ' $r$ ' coalesce to form a large drop of radius ' $R$ ' and volume ' $V$ '. If ' $T$ ' is the surface tension then energy

1

4 VT (\frac{1}{r} - \frac{1}{R})

is released

2

3 VT (\frac{1}{r} - \frac{1}{R})

is released

3 is neither released nor absorbed

4

3 VT (\frac{1}{r} - \frac{1}{R})

is absorbed

Explanation:

B Let the number of small spherical drops be ' $n$ ' and radius 'r'.
After coalesce it forms a big drop having radius ' $R$ '
$∵$ Volume of drops before coalesce will be same after coalesce.
$n \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3}$
${nr}^{3} = R^{3}$
$n = \frac{R^{3}}{r^{3}}$
When the number of drops coalesce the energy is released.
$∴$ Initial energy of drops $(E_{1}) = n \times 4 π r^{2} T$
Final energy of drop $(E_{2}) = 4 π R^{2} T$
$∴$ Energy released $(U) = E_{2} - E_{1}$
$U = (4 π R^{2} \cdot T - n \cdot 4 π r^{2} \cdot T)$
$U = 4 π T (R^{2} - n r^{2})$
Putting the value of ' $n$ ', we get -
$U = 4 π T (R^{2} - \frac{R^{3}}{r^{3}} \cdot r^{2})$
$U = 4 π {TR}^{3} [\frac{1}{R} - \frac{1}{r}]$
$U = 3 \times \frac{4}{3} π R^{3} T (\frac{1}{R} - \frac{1}{r})$
$U = 3 VT [\frac{1}{R} - \frac{1}{r}] (∵ V = \frac{4}{3} π R^{3})$
$U = - 3 VT [\frac{1}{r} - \frac{1}{R}]$
Hence, $U = 3 VT [\frac{1}{r} - \frac{1}{R}]$ is released.

MHT-CET 2020

Mechanical Properties of Fluids

142838 A mercury drop of radius ' $R$ ' is divided into 27 droplets of same size. The radius ' $r$ ' of each droplet is

1

r = \frac{R}{3}

2

r = \frac{R}{9}

3

r = \frac{R}{27}

4

r = 3 R

Explanation:

A The volume of mercury drop before dividing into 27 drops will be same to volume of 27 drops.
$∴ \frac{4}{3} π R^{3} = 27 \times \frac{4}{3} π r^{3}$
$R^{3} = 27 r^{3}$
$R^{3} = (3)^{3} r^{3}$
$R = 3 r$
$r = \frac{R}{3}$

MHT-CET 2020

Mechanical Properties of Fluids

142839 Under isothermal conditions, two soap bubbles of radii ' $r_{1}$ ' and ' $r_{2}$ ' combine to form a single soap bubble of radius ' $R$ '. The surface tension of soap solution is $(P =$ outside pressure $)$

1

P (R^{3} - r_{1}^{3} - r_{2}^{3}) / 4 (r_{1}^{2} + r_{2}^{2} - R^{2})

2

P (r_{1}^{3} - R^{3} + r_{2}^{3}) / 4 (R^{2} - r_{1}^{2} - r_{2}^{2})

3

P (R^{3} + r_{1}^{3} + r_{2}^{3}) / 4 (r_{1}^{2} + r_{2}^{2} - R^{2})

4

P (r_{1}^{3} - r_{2}^{3} + R^{3}) / 4 (r_{1}^{2} + r_{2}^{2} + R^{2})

Explanation:

A Pressure inside the first bubble $= P + \frac{4 T}{r_{1}}$
Pressure inside the second bubble $= P + \frac{4 T}{r_{2}}$
We know that, $P V = n R θ$
Where, $θ =$ absolute temperature
As mass of the air inside the soap bubble is conserved $n_{1} = n_{2} = n$
and under isothermal condition, temperature
$θ_{1} = θ_{2} = θ$
$∵ (P + \frac{4 T}{r_{1}}) \frac{4 π}{3} r_{1}^{3} = nR θ$
$And (P + \frac{4 T}{r_{2}}) \frac{4 π}{3} r_{2}^{3} = nR θ$
$∵ (P + \frac{4 T}{R}) \frac{4 π}{3} R^{3} = (P + \frac{4 T}{r_{1}}) \frac{4 π}{3} r_{1}^{3} + (P + \frac{4 T}{r_{2}}) \frac{4 π}{3} r_{2}^{3}$
$(P + \frac{4 T}{R}) R^{3} = (P + \frac{4 T}{r_{1}}) r_{1}^{3} + (P + \frac{4 T}{r_{2}}) r_{2}^{3}$
On solving the equation, we get-
$T = \frac{P (R^{3} - r_{1}^{3} - r_{2}^{3})}{4 (r_{1}^{2} + r_{2}^{2} - R^{2})}$

MHT-CET 2020

Mechanical Properties of Fluids

142834 When a mercury drop of a radius ' $R$ ', breaks into ' $n$ ' droplets of equal size, the radius ' $r$ ' of each droplets is

1

r = \frac{R}{\sqrt{n}}

2

r = {Rn}^{\frac{1}{3}}

3

r = \frac{R}{n}

4

r = \frac{R}{n^{1 / 3}}

Explanation:

D The volume of $Hg$ drops before breaks and after breaks will be same.
$∴ \frac{4}{3} π R^{3} = n \times \frac{4}{3} π r^{3}$
$R^{3} = {nr}^{3}$
$\Rightarrow {(\frac{R}{r})}^{3} = n$
$\Rightarrow \frac{R}{r} = n^{1 / 3}$
$r = \frac{R}{n^{1 / 3}}$

MHT-CET 2020

Mechanical Properties of Fluids

142836 The work done in blowing a soap bubble of radius ' $R$ ' is ' $W_{1}$ ' at room temperature. Now the soap solution is heated. From the heated solution another soap bubble of radius ' $2 R$ ' is blown and the work done is ' $W_{2}$ ', Then

1

W_{2} = 0

2

W_{2} < 4 W_{1}

3

W_{2} = 4 W_{1}

4

W_{2} = W_{1}

Explanation:

B The work done to blowing a soap bubble of radius ' $R$ ' and surface tension ' $T$ ' is given as-
$W_{1} = 2 \times 4 π R^{2} \times T = 8 π R^{2} T$
Now, if radius is doubled then work done to blowing the soap bubble is-
$W_{2} = 2 \times 4 π (2 R)^{2} \times T^{'}$
$W_{2} = 4 \times 8 π R^{2} \times 4 T^{'}$
Dividing equation (ii) by (i), we get-
$\frac{W_{2}}{W_{1}} = \frac{4 T^{'}}{T}$
But, in the question soap bubbles is heated then it's surface tension is decrease so, $W_{2} < 4 W_{1}$

MHT-CET 2020

Mechanical Properties of Fluids

142837 A fix number of spherical drops of a liquid of radius ' $r$ ' coalesce to form a large drop of radius ' $R$ ' and volume ' $V$ '. If ' $T$ ' is the surface tension then energy

1

4 VT (\frac{1}{r} - \frac{1}{R})

is released

2

3 VT (\frac{1}{r} - \frac{1}{R})

is released

3 is neither released nor absorbed

4

3 VT (\frac{1}{r} - \frac{1}{R})

is absorbed

Explanation:

B Let the number of small spherical drops be ' $n$ ' and radius 'r'.
After coalesce it forms a big drop having radius ' $R$ '
$∵$ Volume of drops before coalesce will be same after coalesce.
$n \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3}$
${nr}^{3} = R^{3}$
$n = \frac{R^{3}}{r^{3}}$
When the number of drops coalesce the energy is released.
$∴$ Initial energy of drops $(E_{1}) = n \times 4 π r^{2} T$
Final energy of drop $(E_{2}) = 4 π R^{2} T$
$∴$ Energy released $(U) = E_{2} - E_{1}$
$U = (4 π R^{2} \cdot T - n \cdot 4 π r^{2} \cdot T)$
$U = 4 π T (R^{2} - n r^{2})$
Putting the value of ' $n$ ', we get -
$U = 4 π T (R^{2} - \frac{R^{3}}{r^{3}} \cdot r^{2})$
$U = 4 π {TR}^{3} [\frac{1}{R} - \frac{1}{r}]$
$U = 3 \times \frac{4}{3} π R^{3} T (\frac{1}{R} - \frac{1}{r})$
$U = 3 VT [\frac{1}{R} - \frac{1}{r}] (∵ V = \frac{4}{3} π R^{3})$
$U = - 3 VT [\frac{1}{r} - \frac{1}{R}]$
Hence, $U = 3 VT [\frac{1}{r} - \frac{1}{R}]$ is released.

MHT-CET 2020

Mechanical Properties of Fluids

142838 A mercury drop of radius ' $R$ ' is divided into 27 droplets of same size. The radius ' $r$ ' of each droplet is

1

r = \frac{R}{3}

2

r = \frac{R}{9}

3

r = \frac{R}{27}

4

r = 3 R

Explanation:

A The volume of mercury drop before dividing into 27 drops will be same to volume of 27 drops.
$∴ \frac{4}{3} π R^{3} = 27 \times \frac{4}{3} π r^{3}$
$R^{3} = 27 r^{3}$
$R^{3} = (3)^{3} r^{3}$
$R = 3 r$
$r = \frac{R}{3}$

MHT-CET 2020

Mechanical Properties of Fluids

142839 Under isothermal conditions, two soap bubbles of radii ' $r_{1}$ ' and ' $r_{2}$ ' combine to form a single soap bubble of radius ' $R$ '. The surface tension of soap solution is $(P =$ outside pressure $)$

1

P (R^{3} - r_{1}^{3} - r_{2}^{3}) / 4 (r_{1}^{2} + r_{2}^{2} - R^{2})

2

P (r_{1}^{3} - R^{3} + r_{2}^{3}) / 4 (R^{2} - r_{1}^{2} - r_{2}^{2})

3

P (R^{3} + r_{1}^{3} + r_{2}^{3}) / 4 (r_{1}^{2} + r_{2}^{2} - R^{2})

4

P (r_{1}^{3} - r_{2}^{3} + R^{3}) / 4 (r_{1}^{2} + r_{2}^{2} + R^{2})

Explanation:

A Pressure inside the first bubble $= P + \frac{4 T}{r_{1}}$
Pressure inside the second bubble $= P + \frac{4 T}{r_{2}}$
We know that, $P V = n R θ$
Where, $θ =$ absolute temperature
As mass of the air inside the soap bubble is conserved $n_{1} = n_{2} = n$
and under isothermal condition, temperature
$θ_{1} = θ_{2} = θ$
$∵ (P + \frac{4 T}{r_{1}}) \frac{4 π}{3} r_{1}^{3} = nR θ$
$And (P + \frac{4 T}{r_{2}}) \frac{4 π}{3} r_{2}^{3} = nR θ$
$∵ (P + \frac{4 T}{R}) \frac{4 π}{3} R^{3} = (P + \frac{4 T}{r_{1}}) \frac{4 π}{3} r_{1}^{3} + (P + \frac{4 T}{r_{2}}) \frac{4 π}{3} r_{2}^{3}$
$(P + \frac{4 T}{R}) R^{3} = (P + \frac{4 T}{r_{1}}) r_{1}^{3} + (P + \frac{4 T}{r_{2}}) r_{2}^{3}$
On solving the equation, we get-
$T = \frac{P (R^{3} - r_{1}^{3} - r_{2}^{3})}{4 (r_{1}^{2} + r_{2}^{2} - R^{2})}$

MHT-CET 2020