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Mechanical Properties of Fluids

143013 Two soap bubbles of radii $3 mm$ and $4 mm$ confined in vacuum coalesce isothermally to form a new bubble. The radius of the bubble formed (in mm) is

1 3

2 3.5

3 4

4 5

5 7

Explanation:

D Let the radii of two soap bubble is $r_{1}$ and $r_{2}$ and radius of coalesced bubble is ' $r$ '
$∵ r_{1} = 3 mm, r_{2} = 4 mm$
Since, a soap bubble has two free surfaces total surface area of the coalesced bubbles-
$A = 2 \times 4 π r^{2} = 8 π r^{2}$
$A = 8 π r^{2}$
Potential energy of the coalesced bubbles
$W = 8 π r^{2} T$
Similarly if $W_{1}$ and $W_{2}$ are the potential energies of the two given bubbles, then-
$W_{1} = 8 π r_{1}^{2} T & W_{2} = 8 π r_{2}^{2} T$
$∵$ The total energy of the system is conserved-
$∴ 8 π r^{2} T = 8 π r_{1}^{2} T + 8 π r_{2}^{2} T$
$r^{2} = r_{1}^{2} + r_{2}^{2}$
$r^{2} = (3)^{2} + (4)^{2}$
$r^{2} = 9 + 16$
$r = \sqrt{25}$
$r = 5 cm$
$8 π T (r)^{2} = 8 π T {(r_{1})}^{2} + 8 π {Tr}_{2}^{2}$

Kerala CEE -2018

Mechanical Properties of Fluids

143014 Two soap bubbles each with radius $r_{1}$ and $r_{2}$ coalesce in vacuum under isothermal conditions to form a bigger bubble of radius $R$ . then, $R$ is equal to

1

\sqrt{r_{1}^{2} + r_{2}^{2}}

2

\sqrt{r_{1}^{2} - r_{2}^{2}}

3

r_{1} + r_{2}

4

\frac{\sqrt{r_{1}^{2} + r_{2}^{2}}}{2}

5

2 \sqrt{r_{1}^{2} + r_{2}^{2}}

Explanation:

A Let, the pressure and volume of given bubbles is $P_{1}, P_{2}$ and $V_{1}, V_{2}$ .
According to question, after coalesce the volume and pressure of bigger bubble is $V$ and $P$ .
Now, According to Boyle's law-
At constant temperature, $P . V = C$
According to question, both bubbles coalesce-
So, $P_{1} V_{1} + P_{2} V_{2} = PV$
$∵$ Excess pressure inside a soap bubble is given as
$P_{1} = \frac{2 T}{r_{1}}$
Where, $T =$ Surface tension, $r_{1} =$ radius of soap bubble
$P_{2} = \frac{2 T}{r_{2}}$
$P = \frac{2 T}{R}$
Volume of soap bubbles -
$V = \frac{4}{3} π R^{3}$
$V_{1} = \frac{4}{3} π R_{1}^{3}$
From equation number (1) we get,
$(\frac{2 T}{r_{1}}) \times \frac{4}{3} π r_{1}^{3} + (\frac{2 T}{r_{2}}) \times \frac{4}{3} π r_{2}^{3} = (\frac{2 T}{R}) \times \frac{4}{3} π R^{3}$
$2 T \times \frac{4 π}{3} (\frac{r_{1}^{3}}{r_{1}} + \frac{r_{2}^{3}}{r_{2}}) = \frac{R^{3}}{R} \times (2 T \times \frac{4 π}{3})$
$(r_{1}^{2} + r_{2}^{2}) = R^{2}$
$R = \sqrt{r_{1}^{2} + r_{2}^{2}}$

CGPET 2005

Mechanical Properties of Fluids

143015 A ball of radius $r$ and density $ρ$ falls freely under gravity through a distance $h$ before entering water. Velocity of ball does not change even on entering water. If viscosity of water is $η$ the value of $h$ is given by

1

\frac{2}{9} r^{2} (\frac{1 - ρ}{η}) g

2

\frac{2}{81} r^{2} (\frac{ρ - 1}{η}) g

3

\frac{2}{81} r^{4} {(\frac{ρ - 1}{η})}^{2} g

4

\frac{2}{9} r^{4} {(\frac{ρ - 1}{η})}^{2} g

Explanation:

C Given that,
Radius of ball $= r$
Density of ball $= ρ$
Viscosity of water $= η$
The velocity of ball when it strikes the water surface
$v = \sqrt{2 gh}$

Now,
Terminal velocity of ball inside the water will be-
$v = \frac{2}{9} \cdot \frac{r^{2} g}{η} \frac{(ρ - 1)}{η}$
According to question-
$\sqrt{2 g h} = \frac{2}{9} \cdot r^{2} g \frac{(ρ - 1)}{η}$
Squaring both side-
$2 gh = \frac{4}{81} \frac{r^{4} g^{2}}{η^{2}} (ρ - 1)^{2}$
$h = \frac{2}{81} r^{4} {(\frac{ρ - 1}{η})}^{2} g$

AIIMS-27.05.2018(M)

Mechanical Properties of Fluids

143016 A spherical solid ball of volume $V$ is made of a material of density $ρ_{1}$ . It is falling through a liquid of density $ρ_{1} (ρ_{2} < ρ_{1})$ . Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed $v$ , i.e., $F_{viscous} = - {k v}_{t}^{2} (k > 0)$ . The terminal speed of the ball is

1

\sqrt{\frac{Vg (ρ_{1} - ρ_{2})}{k}}

2

\frac{Vg ρ_{1}}{k}

3

\sqrt{\frac{Vg ρ_{1}}{k}}

4

\frac{Vg (ρ_{1} - ρ_{2})}{k}

Explanation:

A Given,
Volume of ball $= V$ , Density of ball $= ρ_{1}$
Density of liquid $= ρ_{2}$
Viscous Force $(F_{V}) = - {Kv}^{2}$
Where, $v_{t} =$ Terminal speed

For equilibrium condition-
Net upward force $=$ Net downward force
$F_{V} + ρ_{2} Vg = V ρ_{1} g {∵ F = mg = v ρ_{1} g}$
$Kv v_{t}^{2} + ρ_{2} Vg = V ρ_{1} g$
$K v_{t}^{2} = (ρ_{1} - ρ_{2}) Vg$
$v_{t} = \sqrt{\frac{V (ρ_{1} - ρ_{2}) g}{k}}$

AIIMS-2013

Mechanical Properties of Fluids

143017 If the terminal speed of a sphere of gold (density $= 19.5 kg / m^{3}$ ) is $0.2 m / s$ in a viscous liquid (density $= 1.5 kg / m^{3}$ ), find the terminal speed of a sphere of silver (density $= 10.5$ $kg / m^{3}$ ) of the same size in the same liquid.

1

0.4 m / s

2

0.133 m / s

3

0.1 m / s

4

0.2 m / s

Explanation:

C Given that
$ρ_{Ag} = 10.5 kg / m^{3}$ and $v_{Ag} =$ ?
$ρ_{Au} = 19.5 kg / m^{3}$ and $v_{Au} = 0.2 m / se$
We know that, and density of liquid $σ_{L} = 1.5 kg / m^{3}$
$V_{T} = \frac{2}{9} {gr}^{2} \frac{(ρ - σ)}{η}$
$V_{T} \propto (ρ - σ)$
$\frac{V_{Au}}{VAg} = \frac{(ρ_{G} - σ_{L})}{(ρ_{s} - σ_{L})}$
$\frac{0.2}{V_{Ag}} = \frac{18}{9} = 2$
$V_{Ag} = 0.1 m / s$

AIIMS-2008

Mechanical Properties of Fluids

143013 Two soap bubbles of radii $3 mm$ and $4 mm$ confined in vacuum coalesce isothermally to form a new bubble. The radius of the bubble formed (in mm) is

1 3

2 3.5

3 4

4 5

5 7

Explanation:

D Let the radii of two soap bubble is $r_{1}$ and $r_{2}$ and radius of coalesced bubble is ' $r$ '
$∵ r_{1} = 3 mm, r_{2} = 4 mm$
Since, a soap bubble has two free surfaces total surface area of the coalesced bubbles-
$A = 2 \times 4 π r^{2} = 8 π r^{2}$
$A = 8 π r^{2}$
Potential energy of the coalesced bubbles
$W = 8 π r^{2} T$
Similarly if $W_{1}$ and $W_{2}$ are the potential energies of the two given bubbles, then-
$W_{1} = 8 π r_{1}^{2} T & W_{2} = 8 π r_{2}^{2} T$
$∵$ The total energy of the system is conserved-
$∴ 8 π r^{2} T = 8 π r_{1}^{2} T + 8 π r_{2}^{2} T$
$r^{2} = r_{1}^{2} + r_{2}^{2}$
$r^{2} = (3)^{2} + (4)^{2}$
$r^{2} = 9 + 16$
$r = \sqrt{25}$
$r = 5 cm$
$8 π T (r)^{2} = 8 π T {(r_{1})}^{2} + 8 π {Tr}_{2}^{2}$

Kerala CEE -2018

Mechanical Properties of Fluids

143014 Two soap bubbles each with radius $r_{1}$ and $r_{2}$ coalesce in vacuum under isothermal conditions to form a bigger bubble of radius $R$ . then, $R$ is equal to

1

\sqrt{r_{1}^{2} + r_{2}^{2}}

2

\sqrt{r_{1}^{2} - r_{2}^{2}}

3

r_{1} + r_{2}

4

\frac{\sqrt{r_{1}^{2} + r_{2}^{2}}}{2}

5

2 \sqrt{r_{1}^{2} + r_{2}^{2}}

Explanation:

A Let, the pressure and volume of given bubbles is $P_{1}, P_{2}$ and $V_{1}, V_{2}$ .
According to question, after coalesce the volume and pressure of bigger bubble is $V$ and $P$ .
Now, According to Boyle's law-
At constant temperature, $P . V = C$
According to question, both bubbles coalesce-
So, $P_{1} V_{1} + P_{2} V_{2} = PV$
$∵$ Excess pressure inside a soap bubble is given as
$P_{1} = \frac{2 T}{r_{1}}$
Where, $T =$ Surface tension, $r_{1} =$ radius of soap bubble
$P_{2} = \frac{2 T}{r_{2}}$
$P = \frac{2 T}{R}$
Volume of soap bubbles -
$V = \frac{4}{3} π R^{3}$
$V_{1} = \frac{4}{3} π R_{1}^{3}$
From equation number (1) we get,
$(\frac{2 T}{r_{1}}) \times \frac{4}{3} π r_{1}^{3} + (\frac{2 T}{r_{2}}) \times \frac{4}{3} π r_{2}^{3} = (\frac{2 T}{R}) \times \frac{4}{3} π R^{3}$
$2 T \times \frac{4 π}{3} (\frac{r_{1}^{3}}{r_{1}} + \frac{r_{2}^{3}}{r_{2}}) = \frac{R^{3}}{R} \times (2 T \times \frac{4 π}{3})$
$(r_{1}^{2} + r_{2}^{2}) = R^{2}$
$R = \sqrt{r_{1}^{2} + r_{2}^{2}}$

CGPET 2005

Mechanical Properties of Fluids

143015 A ball of radius $r$ and density $ρ$ falls freely under gravity through a distance $h$ before entering water. Velocity of ball does not change even on entering water. If viscosity of water is $η$ the value of $h$ is given by

1

\frac{2}{9} r^{2} (\frac{1 - ρ}{η}) g

2

\frac{2}{81} r^{2} (\frac{ρ - 1}{η}) g

3

\frac{2}{81} r^{4} {(\frac{ρ - 1}{η})}^{2} g

4

\frac{2}{9} r^{4} {(\frac{ρ - 1}{η})}^{2} g

Explanation:

C Given that,
Radius of ball $= r$
Density of ball $= ρ$
Viscosity of water $= η$
The velocity of ball when it strikes the water surface
$v = \sqrt{2 gh}$

Now,
Terminal velocity of ball inside the water will be-
$v = \frac{2}{9} \cdot \frac{r^{2} g}{η} \frac{(ρ - 1)}{η}$
According to question-
$\sqrt{2 g h} = \frac{2}{9} \cdot r^{2} g \frac{(ρ - 1)}{η}$
Squaring both side-
$2 gh = \frac{4}{81} \frac{r^{4} g^{2}}{η^{2}} (ρ - 1)^{2}$
$h = \frac{2}{81} r^{4} {(\frac{ρ - 1}{η})}^{2} g$

AIIMS-27.05.2018(M)

Mechanical Properties of Fluids

143016 A spherical solid ball of volume $V$ is made of a material of density $ρ_{1}$ . It is falling through a liquid of density $ρ_{1} (ρ_{2} < ρ_{1})$ . Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed $v$ , i.e., $F_{viscous} = - {k v}_{t}^{2} (k > 0)$ . The terminal speed of the ball is

1

\sqrt{\frac{Vg (ρ_{1} - ρ_{2})}{k}}

2

\frac{Vg ρ_{1}}{k}

3

\sqrt{\frac{Vg ρ_{1}}{k}}

4

\frac{Vg (ρ_{1} - ρ_{2})}{k}

Explanation:

A Given,
Volume of ball $= V$ , Density of ball $= ρ_{1}$
Density of liquid $= ρ_{2}$
Viscous Force $(F_{V}) = - {Kv}^{2}$
Where, $v_{t} =$ Terminal speed

For equilibrium condition-
Net upward force $=$ Net downward force
$F_{V} + ρ_{2} Vg = V ρ_{1} g {∵ F = mg = v ρ_{1} g}$
$Kv v_{t}^{2} + ρ_{2} Vg = V ρ_{1} g$
$K v_{t}^{2} = (ρ_{1} - ρ_{2}) Vg$
$v_{t} = \sqrt{\frac{V (ρ_{1} - ρ_{2}) g}{k}}$

AIIMS-2013

Mechanical Properties of Fluids

143017 If the terminal speed of a sphere of gold (density $= 19.5 kg / m^{3}$ ) is $0.2 m / s$ in a viscous liquid (density $= 1.5 kg / m^{3}$ ), find the terminal speed of a sphere of silver (density $= 10.5$ $kg / m^{3}$ ) of the same size in the same liquid.

1

0.4 m / s

2

0.133 m / s

3

0.1 m / s

4

0.2 m / s

Explanation:

C Given that
$ρ_{Ag} = 10.5 kg / m^{3}$ and $v_{Ag} =$ ?
$ρ_{Au} = 19.5 kg / m^{3}$ and $v_{Au} = 0.2 m / se$
We know that, and density of liquid $σ_{L} = 1.5 kg / m^{3}$
$V_{T} = \frac{2}{9} {gr}^{2} \frac{(ρ - σ)}{η}$
$V_{T} \propto (ρ - σ)$
$\frac{V_{Au}}{VAg} = \frac{(ρ_{G} - σ_{L})}{(ρ_{s} - σ_{L})}$
$\frac{0.2}{V_{Ag}} = \frac{18}{9} = 2$
$V_{Ag} = 0.1 m / s$

AIIMS-2008

Mechanical Properties of Fluids

143013 Two soap bubbles of radii $3 mm$ and $4 mm$ confined in vacuum coalesce isothermally to form a new bubble. The radius of the bubble formed (in mm) is

1 3

2 3.5

3 4

4 5

5 7

Explanation:

D Let the radii of two soap bubble is $r_{1}$ and $r_{2}$ and radius of coalesced bubble is ' $r$ '
$∵ r_{1} = 3 mm, r_{2} = 4 mm$
Since, a soap bubble has two free surfaces total surface area of the coalesced bubbles-
$A = 2 \times 4 π r^{2} = 8 π r^{2}$
$A = 8 π r^{2}$
Potential energy of the coalesced bubbles
$W = 8 π r^{2} T$
Similarly if $W_{1}$ and $W_{2}$ are the potential energies of the two given bubbles, then-
$W_{1} = 8 π r_{1}^{2} T & W_{2} = 8 π r_{2}^{2} T$
$∵$ The total energy of the system is conserved-
$∴ 8 π r^{2} T = 8 π r_{1}^{2} T + 8 π r_{2}^{2} T$
$r^{2} = r_{1}^{2} + r_{2}^{2}$
$r^{2} = (3)^{2} + (4)^{2}$
$r^{2} = 9 + 16$
$r = \sqrt{25}$
$r = 5 cm$
$8 π T (r)^{2} = 8 π T {(r_{1})}^{2} + 8 π {Tr}_{2}^{2}$

Kerala CEE -2018

Mechanical Properties of Fluids

143014 Two soap bubbles each with radius $r_{1}$ and $r_{2}$ coalesce in vacuum under isothermal conditions to form a bigger bubble of radius $R$ . then, $R$ is equal to

1

\sqrt{r_{1}^{2} + r_{2}^{2}}

2

\sqrt{r_{1}^{2} - r_{2}^{2}}

3

r_{1} + r_{2}

4

\frac{\sqrt{r_{1}^{2} + r_{2}^{2}}}{2}

5

2 \sqrt{r_{1}^{2} + r_{2}^{2}}

Explanation:

A Let, the pressure and volume of given bubbles is $P_{1}, P_{2}$ and $V_{1}, V_{2}$ .
According to question, after coalesce the volume and pressure of bigger bubble is $V$ and $P$ .
Now, According to Boyle's law-
At constant temperature, $P . V = C$
According to question, both bubbles coalesce-
So, $P_{1} V_{1} + P_{2} V_{2} = PV$
$∵$ Excess pressure inside a soap bubble is given as
$P_{1} = \frac{2 T}{r_{1}}$
Where, $T =$ Surface tension, $r_{1} =$ radius of soap bubble
$P_{2} = \frac{2 T}{r_{2}}$
$P = \frac{2 T}{R}$
Volume of soap bubbles -
$V = \frac{4}{3} π R^{3}$
$V_{1} = \frac{4}{3} π R_{1}^{3}$
From equation number (1) we get,
$(\frac{2 T}{r_{1}}) \times \frac{4}{3} π r_{1}^{3} + (\frac{2 T}{r_{2}}) \times \frac{4}{3} π r_{2}^{3} = (\frac{2 T}{R}) \times \frac{4}{3} π R^{3}$
$2 T \times \frac{4 π}{3} (\frac{r_{1}^{3}}{r_{1}} + \frac{r_{2}^{3}}{r_{2}}) = \frac{R^{3}}{R} \times (2 T \times \frac{4 π}{3})$
$(r_{1}^{2} + r_{2}^{2}) = R^{2}$
$R = \sqrt{r_{1}^{2} + r_{2}^{2}}$

CGPET 2005

Mechanical Properties of Fluids

143015 A ball of radius $r$ and density $ρ$ falls freely under gravity through a distance $h$ before entering water. Velocity of ball does not change even on entering water. If viscosity of water is $η$ the value of $h$ is given by

1

\frac{2}{9} r^{2} (\frac{1 - ρ}{η}) g

2

\frac{2}{81} r^{2} (\frac{ρ - 1}{η}) g

3

\frac{2}{81} r^{4} {(\frac{ρ - 1}{η})}^{2} g

4

\frac{2}{9} r^{4} {(\frac{ρ - 1}{η})}^{2} g

Explanation:

C Given that,
Radius of ball $= r$
Density of ball $= ρ$
Viscosity of water $= η$
The velocity of ball when it strikes the water surface
$v = \sqrt{2 gh}$

Now,
Terminal velocity of ball inside the water will be-
$v = \frac{2}{9} \cdot \frac{r^{2} g}{η} \frac{(ρ - 1)}{η}$
According to question-
$\sqrt{2 g h} = \frac{2}{9} \cdot r^{2} g \frac{(ρ - 1)}{η}$
Squaring both side-
$2 gh = \frac{4}{81} \frac{r^{4} g^{2}}{η^{2}} (ρ - 1)^{2}$
$h = \frac{2}{81} r^{4} {(\frac{ρ - 1}{η})}^{2} g$

AIIMS-27.05.2018(M)

Mechanical Properties of Fluids

143016 A spherical solid ball of volume $V$ is made of a material of density $ρ_{1}$ . It is falling through a liquid of density $ρ_{1} (ρ_{2} < ρ_{1})$ . Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed $v$ , i.e., $F_{viscous} = - {k v}_{t}^{2} (k > 0)$ . The terminal speed of the ball is

1

\sqrt{\frac{Vg (ρ_{1} - ρ_{2})}{k}}

2

\frac{Vg ρ_{1}}{k}

3

\sqrt{\frac{Vg ρ_{1}}{k}}

4

\frac{Vg (ρ_{1} - ρ_{2})}{k}

Explanation:

A Given,
Volume of ball $= V$ , Density of ball $= ρ_{1}$
Density of liquid $= ρ_{2}$
Viscous Force $(F_{V}) = - {Kv}^{2}$
Where, $v_{t} =$ Terminal speed

For equilibrium condition-
Net upward force $=$ Net downward force
$F_{V} + ρ_{2} Vg = V ρ_{1} g {∵ F = mg = v ρ_{1} g}$
$Kv v_{t}^{2} + ρ_{2} Vg = V ρ_{1} g$
$K v_{t}^{2} = (ρ_{1} - ρ_{2}) Vg$
$v_{t} = \sqrt{\frac{V (ρ_{1} - ρ_{2}) g}{k}}$

AIIMS-2013

Mechanical Properties of Fluids

143017 If the terminal speed of a sphere of gold (density $= 19.5 kg / m^{3}$ ) is $0.2 m / s$ in a viscous liquid (density $= 1.5 kg / m^{3}$ ), find the terminal speed of a sphere of silver (density $= 10.5$ $kg / m^{3}$ ) of the same size in the same liquid.

1

0.4 m / s

2

0.133 m / s

3

0.1 m / s

4

0.2 m / s

Explanation:

C Given that
$ρ_{Ag} = 10.5 kg / m^{3}$ and $v_{Ag} =$ ?
$ρ_{Au} = 19.5 kg / m^{3}$ and $v_{Au} = 0.2 m / se$
We know that, and density of liquid $σ_{L} = 1.5 kg / m^{3}$
$V_{T} = \frac{2}{9} {gr}^{2} \frac{(ρ - σ)}{η}$
$V_{T} \propto (ρ - σ)$
$\frac{V_{Au}}{VAg} = \frac{(ρ_{G} - σ_{L})}{(ρ_{s} - σ_{L})}$
$\frac{0.2}{V_{Ag}} = \frac{18}{9} = 2$
$V_{Ag} = 0.1 m / s$

AIIMS-2008

Mechanical Properties of Fluids

143013 Two soap bubbles of radii $3 mm$ and $4 mm$ confined in vacuum coalesce isothermally to form a new bubble. The radius of the bubble formed (in mm) is

1 3

2 3.5

3 4

4 5

5 7

Explanation:

D Let the radii of two soap bubble is $r_{1}$ and $r_{2}$ and radius of coalesced bubble is ' $r$ '
$∵ r_{1} = 3 mm, r_{2} = 4 mm$
Since, a soap bubble has two free surfaces total surface area of the coalesced bubbles-
$A = 2 \times 4 π r^{2} = 8 π r^{2}$
$A = 8 π r^{2}$
Potential energy of the coalesced bubbles
$W = 8 π r^{2} T$
Similarly if $W_{1}$ and $W_{2}$ are the potential energies of the two given bubbles, then-
$W_{1} = 8 π r_{1}^{2} T & W_{2} = 8 π r_{2}^{2} T$
$∵$ The total energy of the system is conserved-
$∴ 8 π r^{2} T = 8 π r_{1}^{2} T + 8 π r_{2}^{2} T$
$r^{2} = r_{1}^{2} + r_{2}^{2}$
$r^{2} = (3)^{2} + (4)^{2}$
$r^{2} = 9 + 16$
$r = \sqrt{25}$
$r = 5 cm$
$8 π T (r)^{2} = 8 π T {(r_{1})}^{2} + 8 π {Tr}_{2}^{2}$

Kerala CEE -2018

Mechanical Properties of Fluids

143014 Two soap bubbles each with radius $r_{1}$ and $r_{2}$ coalesce in vacuum under isothermal conditions to form a bigger bubble of radius $R$ . then, $R$ is equal to

1

\sqrt{r_{1}^{2} + r_{2}^{2}}

2

\sqrt{r_{1}^{2} - r_{2}^{2}}

3

r_{1} + r_{2}

4

\frac{\sqrt{r_{1}^{2} + r_{2}^{2}}}{2}

5

2 \sqrt{r_{1}^{2} + r_{2}^{2}}

Explanation:

A Let, the pressure and volume of given bubbles is $P_{1}, P_{2}$ and $V_{1}, V_{2}$ .
According to question, after coalesce the volume and pressure of bigger bubble is $V$ and $P$ .
Now, According to Boyle's law-
At constant temperature, $P . V = C$
According to question, both bubbles coalesce-
So, $P_{1} V_{1} + P_{2} V_{2} = PV$
$∵$ Excess pressure inside a soap bubble is given as
$P_{1} = \frac{2 T}{r_{1}}$
Where, $T =$ Surface tension, $r_{1} =$ radius of soap bubble
$P_{2} = \frac{2 T}{r_{2}}$
$P = \frac{2 T}{R}$
Volume of soap bubbles -
$V = \frac{4}{3} π R^{3}$
$V_{1} = \frac{4}{3} π R_{1}^{3}$
From equation number (1) we get,
$(\frac{2 T}{r_{1}}) \times \frac{4}{3} π r_{1}^{3} + (\frac{2 T}{r_{2}}) \times \frac{4}{3} π r_{2}^{3} = (\frac{2 T}{R}) \times \frac{4}{3} π R^{3}$
$2 T \times \frac{4 π}{3} (\frac{r_{1}^{3}}{r_{1}} + \frac{r_{2}^{3}}{r_{2}}) = \frac{R^{3}}{R} \times (2 T \times \frac{4 π}{3})$
$(r_{1}^{2} + r_{2}^{2}) = R^{2}$
$R = \sqrt{r_{1}^{2} + r_{2}^{2}}$

CGPET 2005

Mechanical Properties of Fluids

143015 A ball of radius $r$ and density $ρ$ falls freely under gravity through a distance $h$ before entering water. Velocity of ball does not change even on entering water. If viscosity of water is $η$ the value of $h$ is given by

1

\frac{2}{9} r^{2} (\frac{1 - ρ}{η}) g

2

\frac{2}{81} r^{2} (\frac{ρ - 1}{η}) g

3

\frac{2}{81} r^{4} {(\frac{ρ - 1}{η})}^{2} g

4

\frac{2}{9} r^{4} {(\frac{ρ - 1}{η})}^{2} g

Explanation:

C Given that,
Radius of ball $= r$
Density of ball $= ρ$
Viscosity of water $= η$
The velocity of ball when it strikes the water surface
$v = \sqrt{2 gh}$

Now,
Terminal velocity of ball inside the water will be-
$v = \frac{2}{9} \cdot \frac{r^{2} g}{η} \frac{(ρ - 1)}{η}$
According to question-
$\sqrt{2 g h} = \frac{2}{9} \cdot r^{2} g \frac{(ρ - 1)}{η}$
Squaring both side-
$2 gh = \frac{4}{81} \frac{r^{4} g^{2}}{η^{2}} (ρ - 1)^{2}$
$h = \frac{2}{81} r^{4} {(\frac{ρ - 1}{η})}^{2} g$

AIIMS-27.05.2018(M)

Mechanical Properties of Fluids

143016 A spherical solid ball of volume $V$ is made of a material of density $ρ_{1}$ . It is falling through a liquid of density $ρ_{1} (ρ_{2} < ρ_{1})$ . Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed $v$ , i.e., $F_{viscous} = - {k v}_{t}^{2} (k > 0)$ . The terminal speed of the ball is

1

\sqrt{\frac{Vg (ρ_{1} - ρ_{2})}{k}}

2

\frac{Vg ρ_{1}}{k}

3

\sqrt{\frac{Vg ρ_{1}}{k}}

4

\frac{Vg (ρ_{1} - ρ_{2})}{k}

Explanation:

A Given,
Volume of ball $= V$ , Density of ball $= ρ_{1}$
Density of liquid $= ρ_{2}$
Viscous Force $(F_{V}) = - {Kv}^{2}$
Where, $v_{t} =$ Terminal speed

For equilibrium condition-
Net upward force $=$ Net downward force
$F_{V} + ρ_{2} Vg = V ρ_{1} g {∵ F = mg = v ρ_{1} g}$
$Kv v_{t}^{2} + ρ_{2} Vg = V ρ_{1} g$
$K v_{t}^{2} = (ρ_{1} - ρ_{2}) Vg$
$v_{t} = \sqrt{\frac{V (ρ_{1} - ρ_{2}) g}{k}}$

AIIMS-2013

Mechanical Properties of Fluids

143017 If the terminal speed of a sphere of gold (density $= 19.5 kg / m^{3}$ ) is $0.2 m / s$ in a viscous liquid (density $= 1.5 kg / m^{3}$ ), find the terminal speed of a sphere of silver (density $= 10.5$ $kg / m^{3}$ ) of the same size in the same liquid.

1

0.4 m / s

2

0.133 m / s

3

0.1 m / s

4

0.2 m / s

Explanation:

C Given that
$ρ_{Ag} = 10.5 kg / m^{3}$ and $v_{Ag} =$ ?
$ρ_{Au} = 19.5 kg / m^{3}$ and $v_{Au} = 0.2 m / se$
We know that, and density of liquid $σ_{L} = 1.5 kg / m^{3}$
$V_{T} = \frac{2}{9} {gr}^{2} \frac{(ρ - σ)}{η}$
$V_{T} \propto (ρ - σ)$
$\frac{V_{Au}}{VAg} = \frac{(ρ_{G} - σ_{L})}{(ρ_{s} - σ_{L})}$
$\frac{0.2}{V_{Ag}} = \frac{18}{9} = 2$
$V_{Ag} = 0.1 m / s$

AIIMS-2008

Mechanical Properties of Fluids

143013 Two soap bubbles of radii $3 mm$ and $4 mm$ confined in vacuum coalesce isothermally to form a new bubble. The radius of the bubble formed (in mm) is

1 3

2 3.5

3 4

4 5

5 7

Explanation:

D Let the radii of two soap bubble is $r_{1}$ and $r_{2}$ and radius of coalesced bubble is ' $r$ '
$∵ r_{1} = 3 mm, r_{2} = 4 mm$
Since, a soap bubble has two free surfaces total surface area of the coalesced bubbles-
$A = 2 \times 4 π r^{2} = 8 π r^{2}$
$A = 8 π r^{2}$
Potential energy of the coalesced bubbles
$W = 8 π r^{2} T$
Similarly if $W_{1}$ and $W_{2}$ are the potential energies of the two given bubbles, then-
$W_{1} = 8 π r_{1}^{2} T & W_{2} = 8 π r_{2}^{2} T$
$∵$ The total energy of the system is conserved-
$∴ 8 π r^{2} T = 8 π r_{1}^{2} T + 8 π r_{2}^{2} T$
$r^{2} = r_{1}^{2} + r_{2}^{2}$
$r^{2} = (3)^{2} + (4)^{2}$
$r^{2} = 9 + 16$
$r = \sqrt{25}$
$r = 5 cm$
$8 π T (r)^{2} = 8 π T {(r_{1})}^{2} + 8 π {Tr}_{2}^{2}$

Kerala CEE -2018

Mechanical Properties of Fluids

143014 Two soap bubbles each with radius $r_{1}$ and $r_{2}$ coalesce in vacuum under isothermal conditions to form a bigger bubble of radius $R$ . then, $R$ is equal to

1

\sqrt{r_{1}^{2} + r_{2}^{2}}

2

\sqrt{r_{1}^{2} - r_{2}^{2}}

3

r_{1} + r_{2}

4

\frac{\sqrt{r_{1}^{2} + r_{2}^{2}}}{2}

5

2 \sqrt{r_{1}^{2} + r_{2}^{2}}

Explanation:

A Let, the pressure and volume of given bubbles is $P_{1}, P_{2}$ and $V_{1}, V_{2}$ .
According to question, after coalesce the volume and pressure of bigger bubble is $V$ and $P$ .
Now, According to Boyle's law-
At constant temperature, $P . V = C$
According to question, both bubbles coalesce-
So, $P_{1} V_{1} + P_{2} V_{2} = PV$
$∵$ Excess pressure inside a soap bubble is given as
$P_{1} = \frac{2 T}{r_{1}}$
Where, $T =$ Surface tension, $r_{1} =$ radius of soap bubble
$P_{2} = \frac{2 T}{r_{2}}$
$P = \frac{2 T}{R}$
Volume of soap bubbles -
$V = \frac{4}{3} π R^{3}$
$V_{1} = \frac{4}{3} π R_{1}^{3}$
From equation number (1) we get,
$(\frac{2 T}{r_{1}}) \times \frac{4}{3} π r_{1}^{3} + (\frac{2 T}{r_{2}}) \times \frac{4}{3} π r_{2}^{3} = (\frac{2 T}{R}) \times \frac{4}{3} π R^{3}$
$2 T \times \frac{4 π}{3} (\frac{r_{1}^{3}}{r_{1}} + \frac{r_{2}^{3}}{r_{2}}) = \frac{R^{3}}{R} \times (2 T \times \frac{4 π}{3})$
$(r_{1}^{2} + r_{2}^{2}) = R^{2}$
$R = \sqrt{r_{1}^{2} + r_{2}^{2}}$

CGPET 2005

Mechanical Properties of Fluids

143015 A ball of radius $r$ and density $ρ$ falls freely under gravity through a distance $h$ before entering water. Velocity of ball does not change even on entering water. If viscosity of water is $η$ the value of $h$ is given by

1

\frac{2}{9} r^{2} (\frac{1 - ρ}{η}) g

2

\frac{2}{81} r^{2} (\frac{ρ - 1}{η}) g

3

\frac{2}{81} r^{4} {(\frac{ρ - 1}{η})}^{2} g

4

\frac{2}{9} r^{4} {(\frac{ρ - 1}{η})}^{2} g

Explanation:

C Given that,
Radius of ball $= r$
Density of ball $= ρ$
Viscosity of water $= η$
The velocity of ball when it strikes the water surface
$v = \sqrt{2 gh}$

Now,
Terminal velocity of ball inside the water will be-
$v = \frac{2}{9} \cdot \frac{r^{2} g}{η} \frac{(ρ - 1)}{η}$
According to question-
$\sqrt{2 g h} = \frac{2}{9} \cdot r^{2} g \frac{(ρ - 1)}{η}$
Squaring both side-
$2 gh = \frac{4}{81} \frac{r^{4} g^{2}}{η^{2}} (ρ - 1)^{2}$
$h = \frac{2}{81} r^{4} {(\frac{ρ - 1}{η})}^{2} g$

AIIMS-27.05.2018(M)

Mechanical Properties of Fluids

143016 A spherical solid ball of volume $V$ is made of a material of density $ρ_{1}$ . It is falling through a liquid of density $ρ_{1} (ρ_{2} < ρ_{1})$ . Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed $v$ , i.e., $F_{viscous} = - {k v}_{t}^{2} (k > 0)$ . The terminal speed of the ball is

1

\sqrt{\frac{Vg (ρ_{1} - ρ_{2})}{k}}

2

\frac{Vg ρ_{1}}{k}

3

\sqrt{\frac{Vg ρ_{1}}{k}}

4

\frac{Vg (ρ_{1} - ρ_{2})}{k}

Explanation:

A Given,
Volume of ball $= V$ , Density of ball $= ρ_{1}$
Density of liquid $= ρ_{2}$
Viscous Force $(F_{V}) = - {Kv}^{2}$
Where, $v_{t} =$ Terminal speed

For equilibrium condition-
Net upward force $=$ Net downward force
$F_{V} + ρ_{2} Vg = V ρ_{1} g {∵ F = mg = v ρ_{1} g}$
$Kv v_{t}^{2} + ρ_{2} Vg = V ρ_{1} g$
$K v_{t}^{2} = (ρ_{1} - ρ_{2}) Vg$
$v_{t} = \sqrt{\frac{V (ρ_{1} - ρ_{2}) g}{k}}$

AIIMS-2013

Mechanical Properties of Fluids

143017 If the terminal speed of a sphere of gold (density $= 19.5 kg / m^{3}$ ) is $0.2 m / s$ in a viscous liquid (density $= 1.5 kg / m^{3}$ ), find the terminal speed of a sphere of silver (density $= 10.5$ $kg / m^{3}$ ) of the same size in the same liquid.

1

0.4 m / s

2

0.133 m / s

3

0.1 m / s

4

0.2 m / s

Explanation:

C Given that
$ρ_{Ag} = 10.5 kg / m^{3}$ and $v_{Ag} =$ ?
$ρ_{Au} = 19.5 kg / m^{3}$ and $v_{Au} = 0.2 m / se$
We know that, and density of liquid $σ_{L} = 1.5 kg / m^{3}$
$V_{T} = \frac{2}{9} {gr}^{2} \frac{(ρ - σ)}{η}$
$V_{T} \propto (ρ - σ)$
$\frac{V_{Au}}{VAg} = \frac{(ρ_{G} - σ_{L})}{(ρ_{s} - σ_{L})}$
$\frac{0.2}{V_{Ag}} = \frac{18}{9} = 2$
$V_{Ag} = 0.1 m / s$

AIIMS-2008

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