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Mechanical Properties of Fluids

143062 $n$ identical droplets are charged to $v$ volt each.
If they coalesce to form a single large drop, then its potential will be

1

n^{2 / 3} v

2

n^{1 / 3} v

3

nv

4

v / n

Explanation:

A For small droplets-
Let charge on each droplet $= q$
Let radius of small drop $= r$
$∴$ Capacitance $(C) = 4 π ε_{\circ} r$
$∴$ Charge $(q) = Cv$
$(q) = 4 π ε_{\circ} r v$
For large drop-
Let radius of large drop $= R$
$∴ n \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3}$
$R = n^{1 / 3} r$
Capacitance $(C^{'}) = 4 π ε_{\circ} n^{1 / 3} r$
Let charge on large drop $= Q$
$∴ Q = nq$
Putting the value of $q$ from equation (i), we get
$C^{'} v^{'} = nCv$
$4 π ε_{\circ} n^{1 / 3} rv$
$v^{'} = \frac{n}{n^{1 / 3}} v$
$v^{'} = n^{2 / 3} v$
$4 π ε_{\circ} n^{1 / 3} r^{'} = n 4 π ε_{\circ} r v$

MP PMT 2010

Mechanical Properties of Fluids

143063 Under isothermal conditions, two soap bubbles of radii a and $b$ coalesce to form a single bubble of radius $c$ . If the external pressure is $P$ , then surface tension of the bubble is

1

\frac{P (c^{3} - a^{3} + b^{3})}{4 (a^{2} + b^{2} - c^{2})}

2

\frac{P (c^{3} - a^{3} - b^{3})}{4 (a^{2} + b^{2} - c^{2})}

3

\frac{P (c^{2} + a^{2} - b^{2})}{4 (a^{3} + b^{3} - c^{3})}

4

\frac{P (a^{3} + b^{3} - c^{3})}{4 (a^{2} + b^{2} - c^{2})}

Explanation:

B Given, radius of first soap bubble $= a$
Radius of second soap bubble $= b$
Radius of combine soap bubble $= c$
Let $P_{1}$ and $P_{2}$ be the pressure of first and second soap bubble respectively.

From ideal gas equation,
$PV = n nT$
$∴ \frac{P_{1} V_{1}}{{RT}_{1}} + \frac{P_{2} V_{2}}{{RT}_{2}} = \frac{PV}{RT} [∵ T_{1} = T_{2} = T]$
${∵ n_{1} + n_{2} = n = constant}$
$P_{1} V_{1} + P_{2} V_{2} = PV$
$(P + \frac{4 T}{r_{1}}) V_{1} + (P + \frac{4 T}{r_{2}}) V_{2} = (P + \frac{4 T}{r}) V_{3}$
${P + \frac{4 T}{r} is the excess pressure in bubble}$
$(P + \frac{4 T}{r_{1}}) (\frac{4}{3} π r_{1}^{3}) + (P + \frac{4 T}{r_{2}}) (\frac{4}{3} π r_{2}^{3}) = (P + \frac{4 T}{r}) (\frac{4}{3} π r^{3})$
${Volume of sphere = \frac{4}{3} π r^{3}}$
$T = \frac{P (r^{3} - r_{1}^{3} - r_{2}^{3})}{4 (r_{1}^{2} + r_{2}^{2} - r^{2})}$
${∵ r = c, r_{1} = a, r_{2} = b}$
$T = \frac{P (c^{3} - a^{3} - b^{3})}{4 (a^{2} + b^{2} - c^{2})}$

WB JEE 2021

Mechanical Properties of Fluids

143064 A spherical liquid drop is placed on a horizontal plane. A small disturbance causes the volume of the drop to oscillate. The time period of oscillation (T) of the liquid drop depends on radius ( $r$ ) of the drop, density $(ρ)$ and surface tension (s) of the liquid. Which among the following will be a possible expression for $T$ (where, $k$ is a dimensionless constant)?

1

k \sqrt{\frac{ρ r}{s}}

2

k \sqrt{\frac{ρ^{2} r}{s}}

3

k \sqrt{\frac{ρ r^{3}}{s}}

4

k \sqrt{\frac{ρ r^{3}}{s^{2}}}

Explanation:

C Given,
Time period of oscillation of liquid drop $= T$
Radius of liquid drop $= r$
Density of liquid drop $= ρ$
Surface tension of liquid drop $= s$
Let we assume that
$T \propto r^{α}$
$T \propto ρ^{β}$
$T \propto s^{γ}$
Combining (i), (ii) and (iii), we get-
$T \propto r^{α} ρ^{β} s^{γ}$
$T = {kr}^{α} ρ^{β} s^{γ}$
Taking dimensional units on both side, we get,
$[T] = [r]^{α} [ρ]^{β} [s]^{γ}$
$[T] = [L]^{α} {[{ML}^{- 3}]}^{β} {[{MT}^{- 2}]}^{γ}$
$[T] = [M]^{β + γ} [L]^{α - 3 β} [T]^{- 2 γ}$
On comparing we get,
(a) $β + γ = 0$
(b) $α - 3 β = 0$
(c) $- 2 γ = 1 \Rightarrow γ = - \frac{1}{2}$
From (a), $β - \frac{1}{2} = 0 \Rightarrow β = \frac{1}{2}$
From (b), $α - \frac{3}{2} = 0 \Rightarrow α = \frac{3}{2}$
$T = k r r^{3 / 2} ρ^{1 / 2} s^{- 1 / 2}$
$T = k \sqrt{\frac{ρ^{3}}{s}}$

WB JEE 2018

Mechanical Properties of Fluids

143065 1000 droplets of water having $2 mm$ diameter each coalesce to form a single drop. Given the surface tension of water is $0.072 {Nm}^{- 1}$ . The energy loss in the process is

1

8.146 \times 10^{- 4} J

2

4.4 \times 10^{- 4} J

3

2.108 \times 10^{- 5} J

4

4.7 \times 10^{- 1} J

Explanation:

A Given, diameter of water droplet (d) $= 2 mm$ ,
Radius(r) $= 1 mm = 1 \times 10^{- 3} m, T = 0.072 {Nm}^{- 1}, Δ E =$ ?
Let radius of large droplet $= R$
The amount of water must be same before and after the coalesce-
$1000 \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3}$
Decrease in the surface area,
$Δ A = 1000 \times 4 π r^{2} - 4 π R^{2}$
$= 1000 \times 4 π r^{2} - 4 π (10 r)^{2}$
$= 4 π r^{2} (1000 - 100)$
$= 4 π r^{2} \times 900$
Required energy loss in the process, $Δ E = T . Δ A$
$Δ E = 0.072 \times 900 \times 4 π {(1 \times 10^{- 3})}^{2}$
$Δ E = 8.146 \times 10^{- 4} J$

WB JEE 2016

Mechanical Properties of Fluids

143066 A gas bubble of $2 cm$ diameter rises through a liquid of density $1.75 g {cm}^{- 3}$ with a fixed speed of $0.35 {cms}^{- 1}$ . Neglect the density of the gas. The coefficient of viscosity of the liquid is

1 870 poise

2 1120 poise

3 982 poise

4 1089 poise

Explanation:

D Given, radius (r) $= 1 cm$
Density $(ρ) = 1.75 g {cm}^{- 3}$
Speed $(v) = 0.35 cm s^{- 1}$
$g = 980 cm s^{- 2}$
$η =$ ?
As gas bubble is moving upward with constant speed so Buoyant force $=$ Viscous drag force
$F_{b} = F_{v} \Rightarrow (\frac{4}{3} π r^{3}) ρ g = 6 π η r v$
$η = \frac{2 r^{2} ρ g}{9 v}$
$η = \frac{2}{9} \times \frac{1^{2} \times 1.75 \times 980}{0.35}$
$= 1088.8 \approx 1089$
$η = 1089 Poise$

WB JEE 2016

Mechanical Properties of Fluids

143062 $n$ identical droplets are charged to $v$ volt each.
If they coalesce to form a single large drop, then its potential will be

1

n^{2 / 3} v

2

n^{1 / 3} v

3

nv

4

v / n

Explanation:

A For small droplets-
Let charge on each droplet $= q$
Let radius of small drop $= r$
$∴$ Capacitance $(C) = 4 π ε_{\circ} r$
$∴$ Charge $(q) = Cv$
$(q) = 4 π ε_{\circ} r v$
For large drop-
Let radius of large drop $= R$
$∴ n \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3}$
$R = n^{1 / 3} r$
Capacitance $(C^{'}) = 4 π ε_{\circ} n^{1 / 3} r$
Let charge on large drop $= Q$
$∴ Q = nq$
Putting the value of $q$ from equation (i), we get
$C^{'} v^{'} = nCv$
$4 π ε_{\circ} n^{1 / 3} rv$
$v^{'} = \frac{n}{n^{1 / 3}} v$
$v^{'} = n^{2 / 3} v$
$4 π ε_{\circ} n^{1 / 3} r^{'} = n 4 π ε_{\circ} r v$

MP PMT 2010

Mechanical Properties of Fluids

143063 Under isothermal conditions, two soap bubbles of radii a and $b$ coalesce to form a single bubble of radius $c$ . If the external pressure is $P$ , then surface tension of the bubble is

1

\frac{P (c^{3} - a^{3} + b^{3})}{4 (a^{2} + b^{2} - c^{2})}

2

\frac{P (c^{3} - a^{3} - b^{3})}{4 (a^{2} + b^{2} - c^{2})}

3

\frac{P (c^{2} + a^{2} - b^{2})}{4 (a^{3} + b^{3} - c^{3})}

4

\frac{P (a^{3} + b^{3} - c^{3})}{4 (a^{2} + b^{2} - c^{2})}

Explanation:

B Given, radius of first soap bubble $= a$
Radius of second soap bubble $= b$
Radius of combine soap bubble $= c$
Let $P_{1}$ and $P_{2}$ be the pressure of first and second soap bubble respectively.

From ideal gas equation,
$PV = n nT$
$∴ \frac{P_{1} V_{1}}{{RT}_{1}} + \frac{P_{2} V_{2}}{{RT}_{2}} = \frac{PV}{RT} [∵ T_{1} = T_{2} = T]$
${∵ n_{1} + n_{2} = n = constant}$
$P_{1} V_{1} + P_{2} V_{2} = PV$
$(P + \frac{4 T}{r_{1}}) V_{1} + (P + \frac{4 T}{r_{2}}) V_{2} = (P + \frac{4 T}{r}) V_{3}$
${P + \frac{4 T}{r} is the excess pressure in bubble}$
$(P + \frac{4 T}{r_{1}}) (\frac{4}{3} π r_{1}^{3}) + (P + \frac{4 T}{r_{2}}) (\frac{4}{3} π r_{2}^{3}) = (P + \frac{4 T}{r}) (\frac{4}{3} π r^{3})$
${Volume of sphere = \frac{4}{3} π r^{3}}$
$T = \frac{P (r^{3} - r_{1}^{3} - r_{2}^{3})}{4 (r_{1}^{2} + r_{2}^{2} - r^{2})}$
${∵ r = c, r_{1} = a, r_{2} = b}$
$T = \frac{P (c^{3} - a^{3} - b^{3})}{4 (a^{2} + b^{2} - c^{2})}$

WB JEE 2021

Mechanical Properties of Fluids

143064 A spherical liquid drop is placed on a horizontal plane. A small disturbance causes the volume of the drop to oscillate. The time period of oscillation (T) of the liquid drop depends on radius ( $r$ ) of the drop, density $(ρ)$ and surface tension (s) of the liquid. Which among the following will be a possible expression for $T$ (where, $k$ is a dimensionless constant)?

1

k \sqrt{\frac{ρ r}{s}}

2

k \sqrt{\frac{ρ^{2} r}{s}}

3

k \sqrt{\frac{ρ r^{3}}{s}}

4

k \sqrt{\frac{ρ r^{3}}{s^{2}}}

Explanation:

C Given,
Time period of oscillation of liquid drop $= T$
Radius of liquid drop $= r$
Density of liquid drop $= ρ$
Surface tension of liquid drop $= s$
Let we assume that
$T \propto r^{α}$
$T \propto ρ^{β}$
$T \propto s^{γ}$
Combining (i), (ii) and (iii), we get-
$T \propto r^{α} ρ^{β} s^{γ}$
$T = {kr}^{α} ρ^{β} s^{γ}$
Taking dimensional units on both side, we get,
$[T] = [r]^{α} [ρ]^{β} [s]^{γ}$
$[T] = [L]^{α} {[{ML}^{- 3}]}^{β} {[{MT}^{- 2}]}^{γ}$
$[T] = [M]^{β + γ} [L]^{α - 3 β} [T]^{- 2 γ}$
On comparing we get,
(a) $β + γ = 0$
(b) $α - 3 β = 0$
(c) $- 2 γ = 1 \Rightarrow γ = - \frac{1}{2}$
From (a), $β - \frac{1}{2} = 0 \Rightarrow β = \frac{1}{2}$
From (b), $α - \frac{3}{2} = 0 \Rightarrow α = \frac{3}{2}$
$T = k r r^{3 / 2} ρ^{1 / 2} s^{- 1 / 2}$
$T = k \sqrt{\frac{ρ^{3}}{s}}$

WB JEE 2018

Mechanical Properties of Fluids

143065 1000 droplets of water having $2 mm$ diameter each coalesce to form a single drop. Given the surface tension of water is $0.072 {Nm}^{- 1}$ . The energy loss in the process is

1

8.146 \times 10^{- 4} J

2

4.4 \times 10^{- 4} J

3

2.108 \times 10^{- 5} J

4

4.7 \times 10^{- 1} J

Explanation:

A Given, diameter of water droplet (d) $= 2 mm$ ,
Radius(r) $= 1 mm = 1 \times 10^{- 3} m, T = 0.072 {Nm}^{- 1}, Δ E =$ ?
Let radius of large droplet $= R$
The amount of water must be same before and after the coalesce-
$1000 \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3}$
Decrease in the surface area,
$Δ A = 1000 \times 4 π r^{2} - 4 π R^{2}$
$= 1000 \times 4 π r^{2} - 4 π (10 r)^{2}$
$= 4 π r^{2} (1000 - 100)$
$= 4 π r^{2} \times 900$
Required energy loss in the process, $Δ E = T . Δ A$
$Δ E = 0.072 \times 900 \times 4 π {(1 \times 10^{- 3})}^{2}$
$Δ E = 8.146 \times 10^{- 4} J$

WB JEE 2016

Mechanical Properties of Fluids

143066 A gas bubble of $2 cm$ diameter rises through a liquid of density $1.75 g {cm}^{- 3}$ with a fixed speed of $0.35 {cms}^{- 1}$ . Neglect the density of the gas. The coefficient of viscosity of the liquid is

1 870 poise

2 1120 poise

3 982 poise

4 1089 poise

Explanation:

D Given, radius (r) $= 1 cm$
Density $(ρ) = 1.75 g {cm}^{- 3}$
Speed $(v) = 0.35 cm s^{- 1}$
$g = 980 cm s^{- 2}$
$η =$ ?
As gas bubble is moving upward with constant speed so Buoyant force $=$ Viscous drag force
$F_{b} = F_{v} \Rightarrow (\frac{4}{3} π r^{3}) ρ g = 6 π η r v$
$η = \frac{2 r^{2} ρ g}{9 v}$
$η = \frac{2}{9} \times \frac{1^{2} \times 1.75 \times 980}{0.35}$
$= 1088.8 \approx 1089$
$η = 1089 Poise$

WB JEE 2016

Mechanical Properties of Fluids

143062 $n$ identical droplets are charged to $v$ volt each.
If they coalesce to form a single large drop, then its potential will be

1

n^{2 / 3} v

2

n^{1 / 3} v

3

nv

4

v / n

Explanation:

A For small droplets-
Let charge on each droplet $= q$
Let radius of small drop $= r$
$∴$ Capacitance $(C) = 4 π ε_{\circ} r$
$∴$ Charge $(q) = Cv$
$(q) = 4 π ε_{\circ} r v$
For large drop-
Let radius of large drop $= R$
$∴ n \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3}$
$R = n^{1 / 3} r$
Capacitance $(C^{'}) = 4 π ε_{\circ} n^{1 / 3} r$
Let charge on large drop $= Q$
$∴ Q = nq$
Putting the value of $q$ from equation (i), we get
$C^{'} v^{'} = nCv$
$4 π ε_{\circ} n^{1 / 3} rv$
$v^{'} = \frac{n}{n^{1 / 3}} v$
$v^{'} = n^{2 / 3} v$
$4 π ε_{\circ} n^{1 / 3} r^{'} = n 4 π ε_{\circ} r v$

MP PMT 2010

Mechanical Properties of Fluids

143063 Under isothermal conditions, two soap bubbles of radii a and $b$ coalesce to form a single bubble of radius $c$ . If the external pressure is $P$ , then surface tension of the bubble is

1

\frac{P (c^{3} - a^{3} + b^{3})}{4 (a^{2} + b^{2} - c^{2})}

2

\frac{P (c^{3} - a^{3} - b^{3})}{4 (a^{2} + b^{2} - c^{2})}

3

\frac{P (c^{2} + a^{2} - b^{2})}{4 (a^{3} + b^{3} - c^{3})}

4

\frac{P (a^{3} + b^{3} - c^{3})}{4 (a^{2} + b^{2} - c^{2})}

Explanation:

B Given, radius of first soap bubble $= a$
Radius of second soap bubble $= b$
Radius of combine soap bubble $= c$
Let $P_{1}$ and $P_{2}$ be the pressure of first and second soap bubble respectively.

From ideal gas equation,
$PV = n nT$
$∴ \frac{P_{1} V_{1}}{{RT}_{1}} + \frac{P_{2} V_{2}}{{RT}_{2}} = \frac{PV}{RT} [∵ T_{1} = T_{2} = T]$
${∵ n_{1} + n_{2} = n = constant}$
$P_{1} V_{1} + P_{2} V_{2} = PV$
$(P + \frac{4 T}{r_{1}}) V_{1} + (P + \frac{4 T}{r_{2}}) V_{2} = (P + \frac{4 T}{r}) V_{3}$
${P + \frac{4 T}{r} is the excess pressure in bubble}$
$(P + \frac{4 T}{r_{1}}) (\frac{4}{3} π r_{1}^{3}) + (P + \frac{4 T}{r_{2}}) (\frac{4}{3} π r_{2}^{3}) = (P + \frac{4 T}{r}) (\frac{4}{3} π r^{3})$
${Volume of sphere = \frac{4}{3} π r^{3}}$
$T = \frac{P (r^{3} - r_{1}^{3} - r_{2}^{3})}{4 (r_{1}^{2} + r_{2}^{2} - r^{2})}$
${∵ r = c, r_{1} = a, r_{2} = b}$
$T = \frac{P (c^{3} - a^{3} - b^{3})}{4 (a^{2} + b^{2} - c^{2})}$

WB JEE 2021

Mechanical Properties of Fluids

143064 A spherical liquid drop is placed on a horizontal plane. A small disturbance causes the volume of the drop to oscillate. The time period of oscillation (T) of the liquid drop depends on radius ( $r$ ) of the drop, density $(ρ)$ and surface tension (s) of the liquid. Which among the following will be a possible expression for $T$ (where, $k$ is a dimensionless constant)?

1

k \sqrt{\frac{ρ r}{s}}

2

k \sqrt{\frac{ρ^{2} r}{s}}

3

k \sqrt{\frac{ρ r^{3}}{s}}

4

k \sqrt{\frac{ρ r^{3}}{s^{2}}}

Explanation:

C Given,
Time period of oscillation of liquid drop $= T$
Radius of liquid drop $= r$
Density of liquid drop $= ρ$
Surface tension of liquid drop $= s$
Let we assume that
$T \propto r^{α}$
$T \propto ρ^{β}$
$T \propto s^{γ}$
Combining (i), (ii) and (iii), we get-
$T \propto r^{α} ρ^{β} s^{γ}$
$T = {kr}^{α} ρ^{β} s^{γ}$
Taking dimensional units on both side, we get,
$[T] = [r]^{α} [ρ]^{β} [s]^{γ}$
$[T] = [L]^{α} {[{ML}^{- 3}]}^{β} {[{MT}^{- 2}]}^{γ}$
$[T] = [M]^{β + γ} [L]^{α - 3 β} [T]^{- 2 γ}$
On comparing we get,
(a) $β + γ = 0$
(b) $α - 3 β = 0$
(c) $- 2 γ = 1 \Rightarrow γ = - \frac{1}{2}$
From (a), $β - \frac{1}{2} = 0 \Rightarrow β = \frac{1}{2}$
From (b), $α - \frac{3}{2} = 0 \Rightarrow α = \frac{3}{2}$
$T = k r r^{3 / 2} ρ^{1 / 2} s^{- 1 / 2}$
$T = k \sqrt{\frac{ρ^{3}}{s}}$

WB JEE 2018

Mechanical Properties of Fluids

143065 1000 droplets of water having $2 mm$ diameter each coalesce to form a single drop. Given the surface tension of water is $0.072 {Nm}^{- 1}$ . The energy loss in the process is

1

8.146 \times 10^{- 4} J

2

4.4 \times 10^{- 4} J

3

2.108 \times 10^{- 5} J

4

4.7 \times 10^{- 1} J

Explanation:

A Given, diameter of water droplet (d) $= 2 mm$ ,
Radius(r) $= 1 mm = 1 \times 10^{- 3} m, T = 0.072 {Nm}^{- 1}, Δ E =$ ?
Let radius of large droplet $= R$
The amount of water must be same before and after the coalesce-
$1000 \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3}$
Decrease in the surface area,
$Δ A = 1000 \times 4 π r^{2} - 4 π R^{2}$
$= 1000 \times 4 π r^{2} - 4 π (10 r)^{2}$
$= 4 π r^{2} (1000 - 100)$
$= 4 π r^{2} \times 900$
Required energy loss in the process, $Δ E = T . Δ A$
$Δ E = 0.072 \times 900 \times 4 π {(1 \times 10^{- 3})}^{2}$
$Δ E = 8.146 \times 10^{- 4} J$

WB JEE 2016

Mechanical Properties of Fluids

143066 A gas bubble of $2 cm$ diameter rises through a liquid of density $1.75 g {cm}^{- 3}$ with a fixed speed of $0.35 {cms}^{- 1}$ . Neglect the density of the gas. The coefficient of viscosity of the liquid is

1 870 poise

2 1120 poise

3 982 poise

4 1089 poise

Explanation:

D Given, radius (r) $= 1 cm$
Density $(ρ) = 1.75 g {cm}^{- 3}$
Speed $(v) = 0.35 cm s^{- 1}$
$g = 980 cm s^{- 2}$
$η =$ ?
As gas bubble is moving upward with constant speed so Buoyant force $=$ Viscous drag force
$F_{b} = F_{v} \Rightarrow (\frac{4}{3} π r^{3}) ρ g = 6 π η r v$
$η = \frac{2 r^{2} ρ g}{9 v}$
$η = \frac{2}{9} \times \frac{1^{2} \times 1.75 \times 980}{0.35}$
$= 1088.8 \approx 1089$
$η = 1089 Poise$

WB JEE 2016

Mechanical Properties of Fluids

143062 $n$ identical droplets are charged to $v$ volt each.
If they coalesce to form a single large drop, then its potential will be

1

n^{2 / 3} v

2

n^{1 / 3} v

3

nv

4

v / n

Explanation:

A For small droplets-
Let charge on each droplet $= q$
Let radius of small drop $= r$
$∴$ Capacitance $(C) = 4 π ε_{\circ} r$
$∴$ Charge $(q) = Cv$
$(q) = 4 π ε_{\circ} r v$
For large drop-
Let radius of large drop $= R$
$∴ n \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3}$
$R = n^{1 / 3} r$
Capacitance $(C^{'}) = 4 π ε_{\circ} n^{1 / 3} r$
Let charge on large drop $= Q$
$∴ Q = nq$
Putting the value of $q$ from equation (i), we get
$C^{'} v^{'} = nCv$
$4 π ε_{\circ} n^{1 / 3} rv$
$v^{'} = \frac{n}{n^{1 / 3}} v$
$v^{'} = n^{2 / 3} v$
$4 π ε_{\circ} n^{1 / 3} r^{'} = n 4 π ε_{\circ} r v$

MP PMT 2010

Mechanical Properties of Fluids

143063 Under isothermal conditions, two soap bubbles of radii a and $b$ coalesce to form a single bubble of radius $c$ . If the external pressure is $P$ , then surface tension of the bubble is

1

\frac{P (c^{3} - a^{3} + b^{3})}{4 (a^{2} + b^{2} - c^{2})}

2

\frac{P (c^{3} - a^{3} - b^{3})}{4 (a^{2} + b^{2} - c^{2})}

3

\frac{P (c^{2} + a^{2} - b^{2})}{4 (a^{3} + b^{3} - c^{3})}

4

\frac{P (a^{3} + b^{3} - c^{3})}{4 (a^{2} + b^{2} - c^{2})}

Explanation:

B Given, radius of first soap bubble $= a$
Radius of second soap bubble $= b$
Radius of combine soap bubble $= c$
Let $P_{1}$ and $P_{2}$ be the pressure of first and second soap bubble respectively.

From ideal gas equation,
$PV = n nT$
$∴ \frac{P_{1} V_{1}}{{RT}_{1}} + \frac{P_{2} V_{2}}{{RT}_{2}} = \frac{PV}{RT} [∵ T_{1} = T_{2} = T]$
${∵ n_{1} + n_{2} = n = constant}$
$P_{1} V_{1} + P_{2} V_{2} = PV$
$(P + \frac{4 T}{r_{1}}) V_{1} + (P + \frac{4 T}{r_{2}}) V_{2} = (P + \frac{4 T}{r}) V_{3}$
${P + \frac{4 T}{r} is the excess pressure in bubble}$
$(P + \frac{4 T}{r_{1}}) (\frac{4}{3} π r_{1}^{3}) + (P + \frac{4 T}{r_{2}}) (\frac{4}{3} π r_{2}^{3}) = (P + \frac{4 T}{r}) (\frac{4}{3} π r^{3})$
${Volume of sphere = \frac{4}{3} π r^{3}}$
$T = \frac{P (r^{3} - r_{1}^{3} - r_{2}^{3})}{4 (r_{1}^{2} + r_{2}^{2} - r^{2})}$
${∵ r = c, r_{1} = a, r_{2} = b}$
$T = \frac{P (c^{3} - a^{3} - b^{3})}{4 (a^{2} + b^{2} - c^{2})}$

WB JEE 2021

Mechanical Properties of Fluids

143064 A spherical liquid drop is placed on a horizontal plane. A small disturbance causes the volume of the drop to oscillate. The time period of oscillation (T) of the liquid drop depends on radius ( $r$ ) of the drop, density $(ρ)$ and surface tension (s) of the liquid. Which among the following will be a possible expression for $T$ (where, $k$ is a dimensionless constant)?

1

k \sqrt{\frac{ρ r}{s}}

2

k \sqrt{\frac{ρ^{2} r}{s}}

3

k \sqrt{\frac{ρ r^{3}}{s}}

4

k \sqrt{\frac{ρ r^{3}}{s^{2}}}

Explanation:

C Given,
Time period of oscillation of liquid drop $= T$
Radius of liquid drop $= r$
Density of liquid drop $= ρ$
Surface tension of liquid drop $= s$
Let we assume that
$T \propto r^{α}$
$T \propto ρ^{β}$
$T \propto s^{γ}$
Combining (i), (ii) and (iii), we get-
$T \propto r^{α} ρ^{β} s^{γ}$
$T = {kr}^{α} ρ^{β} s^{γ}$
Taking dimensional units on both side, we get,
$[T] = [r]^{α} [ρ]^{β} [s]^{γ}$
$[T] = [L]^{α} {[{ML}^{- 3}]}^{β} {[{MT}^{- 2}]}^{γ}$
$[T] = [M]^{β + γ} [L]^{α - 3 β} [T]^{- 2 γ}$
On comparing we get,
(a) $β + γ = 0$
(b) $α - 3 β = 0$
(c) $- 2 γ = 1 \Rightarrow γ = - \frac{1}{2}$
From (a), $β - \frac{1}{2} = 0 \Rightarrow β = \frac{1}{2}$
From (b), $α - \frac{3}{2} = 0 \Rightarrow α = \frac{3}{2}$
$T = k r r^{3 / 2} ρ^{1 / 2} s^{- 1 / 2}$
$T = k \sqrt{\frac{ρ^{3}}{s}}$

WB JEE 2018

Mechanical Properties of Fluids

143065 1000 droplets of water having $2 mm$ diameter each coalesce to form a single drop. Given the surface tension of water is $0.072 {Nm}^{- 1}$ . The energy loss in the process is

1

8.146 \times 10^{- 4} J

2

4.4 \times 10^{- 4} J

3

2.108 \times 10^{- 5} J

4

4.7 \times 10^{- 1} J

Explanation:

A Given, diameter of water droplet (d) $= 2 mm$ ,
Radius(r) $= 1 mm = 1 \times 10^{- 3} m, T = 0.072 {Nm}^{- 1}, Δ E =$ ?
Let radius of large droplet $= R$
The amount of water must be same before and after the coalesce-
$1000 \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3}$
Decrease in the surface area,
$Δ A = 1000 \times 4 π r^{2} - 4 π R^{2}$
$= 1000 \times 4 π r^{2} - 4 π (10 r)^{2}$
$= 4 π r^{2} (1000 - 100)$
$= 4 π r^{2} \times 900$
Required energy loss in the process, $Δ E = T . Δ A$
$Δ E = 0.072 \times 900 \times 4 π {(1 \times 10^{- 3})}^{2}$
$Δ E = 8.146 \times 10^{- 4} J$

WB JEE 2016

Mechanical Properties of Fluids

143066 A gas bubble of $2 cm$ diameter rises through a liquid of density $1.75 g {cm}^{- 3}$ with a fixed speed of $0.35 {cms}^{- 1}$ . Neglect the density of the gas. The coefficient of viscosity of the liquid is

1 870 poise

2 1120 poise

3 982 poise

4 1089 poise

Explanation:

D Given, radius (r) $= 1 cm$
Density $(ρ) = 1.75 g {cm}^{- 3}$
Speed $(v) = 0.35 cm s^{- 1}$
$g = 980 cm s^{- 2}$
$η =$ ?
As gas bubble is moving upward with constant speed so Buoyant force $=$ Viscous drag force
$F_{b} = F_{v} \Rightarrow (\frac{4}{3} π r^{3}) ρ g = 6 π η r v$
$η = \frac{2 r^{2} ρ g}{9 v}$
$η = \frac{2}{9} \times \frac{1^{2} \times 1.75 \times 980}{0.35}$
$= 1088.8 \approx 1089$
$η = 1089 Poise$

WB JEE 2016

Mechanical Properties of Fluids

143062 $n$ identical droplets are charged to $v$ volt each.
If they coalesce to form a single large drop, then its potential will be

1

n^{2 / 3} v

2

n^{1 / 3} v

3

nv

4

v / n

Explanation:

A For small droplets-
Let charge on each droplet $= q$
Let radius of small drop $= r$
$∴$ Capacitance $(C) = 4 π ε_{\circ} r$
$∴$ Charge $(q) = Cv$
$(q) = 4 π ε_{\circ} r v$
For large drop-
Let radius of large drop $= R$
$∴ n \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3}$
$R = n^{1 / 3} r$
Capacitance $(C^{'}) = 4 π ε_{\circ} n^{1 / 3} r$
Let charge on large drop $= Q$
$∴ Q = nq$
Putting the value of $q$ from equation (i), we get
$C^{'} v^{'} = nCv$
$4 π ε_{\circ} n^{1 / 3} rv$
$v^{'} = \frac{n}{n^{1 / 3}} v$
$v^{'} = n^{2 / 3} v$
$4 π ε_{\circ} n^{1 / 3} r^{'} = n 4 π ε_{\circ} r v$

MP PMT 2010

Mechanical Properties of Fluids

143063 Under isothermal conditions, two soap bubbles of radii a and $b$ coalesce to form a single bubble of radius $c$ . If the external pressure is $P$ , then surface tension of the bubble is

1

\frac{P (c^{3} - a^{3} + b^{3})}{4 (a^{2} + b^{2} - c^{2})}

2

\frac{P (c^{3} - a^{3} - b^{3})}{4 (a^{2} + b^{2} - c^{2})}

3

\frac{P (c^{2} + a^{2} - b^{2})}{4 (a^{3} + b^{3} - c^{3})}

4

\frac{P (a^{3} + b^{3} - c^{3})}{4 (a^{2} + b^{2} - c^{2})}

Explanation:

B Given, radius of first soap bubble $= a$
Radius of second soap bubble $= b$
Radius of combine soap bubble $= c$
Let $P_{1}$ and $P_{2}$ be the pressure of first and second soap bubble respectively.

From ideal gas equation,
$PV = n nT$
$∴ \frac{P_{1} V_{1}}{{RT}_{1}} + \frac{P_{2} V_{2}}{{RT}_{2}} = \frac{PV}{RT} [∵ T_{1} = T_{2} = T]$
${∵ n_{1} + n_{2} = n = constant}$
$P_{1} V_{1} + P_{2} V_{2} = PV$
$(P + \frac{4 T}{r_{1}}) V_{1} + (P + \frac{4 T}{r_{2}}) V_{2} = (P + \frac{4 T}{r}) V_{3}$
${P + \frac{4 T}{r} is the excess pressure in bubble}$
$(P + \frac{4 T}{r_{1}}) (\frac{4}{3} π r_{1}^{3}) + (P + \frac{4 T}{r_{2}}) (\frac{4}{3} π r_{2}^{3}) = (P + \frac{4 T}{r}) (\frac{4}{3} π r^{3})$
${Volume of sphere = \frac{4}{3} π r^{3}}$
$T = \frac{P (r^{3} - r_{1}^{3} - r_{2}^{3})}{4 (r_{1}^{2} + r_{2}^{2} - r^{2})}$
${∵ r = c, r_{1} = a, r_{2} = b}$
$T = \frac{P (c^{3} - a^{3} - b^{3})}{4 (a^{2} + b^{2} - c^{2})}$

WB JEE 2021

Mechanical Properties of Fluids

143064 A spherical liquid drop is placed on a horizontal plane. A small disturbance causes the volume of the drop to oscillate. The time period of oscillation (T) of the liquid drop depends on radius ( $r$ ) of the drop, density $(ρ)$ and surface tension (s) of the liquid. Which among the following will be a possible expression for $T$ (where, $k$ is a dimensionless constant)?

1

k \sqrt{\frac{ρ r}{s}}

2

k \sqrt{\frac{ρ^{2} r}{s}}

3

k \sqrt{\frac{ρ r^{3}}{s}}

4

k \sqrt{\frac{ρ r^{3}}{s^{2}}}

Explanation:

C Given,
Time period of oscillation of liquid drop $= T$
Radius of liquid drop $= r$
Density of liquid drop $= ρ$
Surface tension of liquid drop $= s$
Let we assume that
$T \propto r^{α}$
$T \propto ρ^{β}$
$T \propto s^{γ}$
Combining (i), (ii) and (iii), we get-
$T \propto r^{α} ρ^{β} s^{γ}$
$T = {kr}^{α} ρ^{β} s^{γ}$
Taking dimensional units on both side, we get,
$[T] = [r]^{α} [ρ]^{β} [s]^{γ}$
$[T] = [L]^{α} {[{ML}^{- 3}]}^{β} {[{MT}^{- 2}]}^{γ}$
$[T] = [M]^{β + γ} [L]^{α - 3 β} [T]^{- 2 γ}$
On comparing we get,
(a) $β + γ = 0$
(b) $α - 3 β = 0$
(c) $- 2 γ = 1 \Rightarrow γ = - \frac{1}{2}$
From (a), $β - \frac{1}{2} = 0 \Rightarrow β = \frac{1}{2}$
From (b), $α - \frac{3}{2} = 0 \Rightarrow α = \frac{3}{2}$
$T = k r r^{3 / 2} ρ^{1 / 2} s^{- 1 / 2}$
$T = k \sqrt{\frac{ρ^{3}}{s}}$

WB JEE 2018

Mechanical Properties of Fluids

143065 1000 droplets of water having $2 mm$ diameter each coalesce to form a single drop. Given the surface tension of water is $0.072 {Nm}^{- 1}$ . The energy loss in the process is

1

8.146 \times 10^{- 4} J

2

4.4 \times 10^{- 4} J

3

2.108 \times 10^{- 5} J

4

4.7 \times 10^{- 1} J

Explanation:

A Given, diameter of water droplet (d) $= 2 mm$ ,
Radius(r) $= 1 mm = 1 \times 10^{- 3} m, T = 0.072 {Nm}^{- 1}, Δ E =$ ?
Let radius of large droplet $= R$
The amount of water must be same before and after the coalesce-
$1000 \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3}$
Decrease in the surface area,
$Δ A = 1000 \times 4 π r^{2} - 4 π R^{2}$
$= 1000 \times 4 π r^{2} - 4 π (10 r)^{2}$
$= 4 π r^{2} (1000 - 100)$
$= 4 π r^{2} \times 900$
Required energy loss in the process, $Δ E = T . Δ A$
$Δ E = 0.072 \times 900 \times 4 π {(1 \times 10^{- 3})}^{2}$
$Δ E = 8.146 \times 10^{- 4} J$

WB JEE 2016

Mechanical Properties of Fluids

143066 A gas bubble of $2 cm$ diameter rises through a liquid of density $1.75 g {cm}^{- 3}$ with a fixed speed of $0.35 {cms}^{- 1}$ . Neglect the density of the gas. The coefficient of viscosity of the liquid is

1 870 poise

2 1120 poise

3 982 poise

4 1089 poise

Explanation:

D Given, radius (r) $= 1 cm$
Density $(ρ) = 1.75 g {cm}^{- 3}$
Speed $(v) = 0.35 cm s^{- 1}$
$g = 980 cm s^{- 2}$
$η =$ ?
As gas bubble is moving upward with constant speed so Buoyant force $=$ Viscous drag force
$F_{b} = F_{v} \Rightarrow (\frac{4}{3} π r^{3}) ρ g = 6 π η r v$
$η = \frac{2 r^{2} ρ g}{9 v}$
$η = \frac{2}{9} \times \frac{1^{2} \times 1.75 \times 980}{0.35}$
$= 1088.8 \approx 1089$
$η = 1089 Poise$

WB JEE 2016

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