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PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361404 A rain drop of radius $0.3 m m$ has a terminal velocity in air $1 m s^{- 1} .$ The viscosity of air is $18 \times 10^{- 5}$ poise. Find the viscous force on the rain drops.

1

2.05 \times 10^{- 7} N

2

1.018 \times 10^{- 7} N

3

1.05 \times 10^{- 7} N

4

2.058 \times 10^{- 7} N

Explanation:

Here,
$r = 0.3 m m = 0.3 \times 10^{- 3} m, v = 1 m s^{- 1}$
$η = 18 \times 10^{- 5}$ poise
$= 18 \times 10^{- 6}$ deca poise
Viscous force, $F = 6 π η r v$
$= 6 \times \frac{22}{7} \times (18 \times 10^{- 6}) \times (0.3 \times 10^{- 3}) \times 1$
$= 1.018 \times 10^{- 7} N$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361405 Spherical balls of radius ' $R$ ' are falling in a viscous fluid of viscosity ' $η$ ' with a velocity ' $v$ '. The retarding viscous force acting on the spherical ball is

1 Inversely proportional to both radius '

R

' and velocity '

v

'

2 Directly proportional to both radius '

R

' and velocity '

v

'

3 Directly proportional to '

R

' but inversely proportional to '

v

'

4 Inversely proportional to '

R

' but Directly proportional to velocity '

v

'

Explanation:

From Stoke's law
Viscous force $F = 6 π η R v$
hence $F$ is directly proportional to radius & velocity.

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361406 When a ball is released from rest in a very long column of viscous liquid, its downward acceleration is ' $a$ ' (just after release). Then its acceleration when it has acquired two third of the maximum velocity:

1

\frac{a}{3}

2

\frac{2 a}{3}

3

\frac{a}{6}

4 None of these

Explanation:

When the ball is just released, the net force on ball is $m a = m g - F_{b}$
The terminal velocity ' $v_{t}$ ' of the ball is attained when net force on the ball is zero.
$m g = F_{b} + 6 π η r v_{t}$
When the ball acquires $\frac{2}{3} rd$ of its maximum velocity $v_{f}$
$\begin{aligned} m a^{'} = m g - F_{b} - 6 π η r (\frac{2 v_{t}}{3}) \\ m a^{'} = m a - \frac{2}{3} (m a) \\ a^{'} = \frac{a}{3} \end{aligned}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361407 Eight drops of water, each of radius $2 m m$ are falling through air at a terminal velocity of $8 c m s^{- 1}$ . If they coalesce to form a single drop, then the terminal velocity of combined drop will be

1

30 c m s^{- 1}

2

28 c m s^{- 1}

3

24 c m s^{- 1}

4

32 c m s^{- 1}

Explanation:

Let the radius of bigger drop is $R$ and that of smaller drop is $r$ then
$\frac{4}{3} π R^{3} = 8 \times \frac{4}{3} \times π r^{3}$
$or R = 2 r (1)$
Terminal velocity, $v \propto r^{2}$
$∴ \frac{v^{'}}{v} = \frac{R^{2}}{r^{2}} = {(\frac{2 r}{r})}^{2} = 4$
$\Rightarrow v^{'} = 4 v = 4 \times 8 = 32 c m s^{- 1}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361404 A rain drop of radius $0.3 m m$ has a terminal velocity in air $1 m s^{- 1} .$ The viscosity of air is $18 \times 10^{- 5}$ poise. Find the viscous force on the rain drops.

1

2.05 \times 10^{- 7} N

2

1.018 \times 10^{- 7} N

3

1.05 \times 10^{- 7} N

4

2.058 \times 10^{- 7} N

Explanation:

Here,
$r = 0.3 m m = 0.3 \times 10^{- 3} m, v = 1 m s^{- 1}$
$η = 18 \times 10^{- 5}$ poise
$= 18 \times 10^{- 6}$ deca poise
Viscous force, $F = 6 π η r v$
$= 6 \times \frac{22}{7} \times (18 \times 10^{- 6}) \times (0.3 \times 10^{- 3}) \times 1$
$= 1.018 \times 10^{- 7} N$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361405 Spherical balls of radius ' $R$ ' are falling in a viscous fluid of viscosity ' $η$ ' with a velocity ' $v$ '. The retarding viscous force acting on the spherical ball is

1 Inversely proportional to both radius '

R

' and velocity '

v

'

2 Directly proportional to both radius '

R

' and velocity '

v

'

3 Directly proportional to '

R

' but inversely proportional to '

v

'

4 Inversely proportional to '

R

' but Directly proportional to velocity '

v

'

Explanation:

From Stoke's law
Viscous force $F = 6 π η R v$
hence $F$ is directly proportional to radius & velocity.

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361406 When a ball is released from rest in a very long column of viscous liquid, its downward acceleration is ' $a$ ' (just after release). Then its acceleration when it has acquired two third of the maximum velocity:

1

\frac{a}{3}

2

\frac{2 a}{3}

3

\frac{a}{6}

4 None of these

Explanation:

When the ball is just released, the net force on ball is $m a = m g - F_{b}$
The terminal velocity ' $v_{t}$ ' of the ball is attained when net force on the ball is zero.
$m g = F_{b} + 6 π η r v_{t}$
When the ball acquires $\frac{2}{3} rd$ of its maximum velocity $v_{f}$
$\begin{aligned} m a^{'} = m g - F_{b} - 6 π η r (\frac{2 v_{t}}{3}) \\ m a^{'} = m a - \frac{2}{3} (m a) \\ a^{'} = \frac{a}{3} \end{aligned}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361407 Eight drops of water, each of radius $2 m m$ are falling through air at a terminal velocity of $8 c m s^{- 1}$ . If they coalesce to form a single drop, then the terminal velocity of combined drop will be

1

30 c m s^{- 1}

2

28 c m s^{- 1}

3

24 c m s^{- 1}

4

32 c m s^{- 1}

Explanation:

Let the radius of bigger drop is $R$ and that of smaller drop is $r$ then
$\frac{4}{3} π R^{3} = 8 \times \frac{4}{3} \times π r^{3}$
$or R = 2 r (1)$
Terminal velocity, $v \propto r^{2}$
$∴ \frac{v^{'}}{v} = \frac{R^{2}}{r^{2}} = {(\frac{2 r}{r})}^{2} = 4$
$\Rightarrow v^{'} = 4 v = 4 \times 8 = 32 c m s^{- 1}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361404 A rain drop of radius $0.3 m m$ has a terminal velocity in air $1 m s^{- 1} .$ The viscosity of air is $18 \times 10^{- 5}$ poise. Find the viscous force on the rain drops.

1

2.05 \times 10^{- 7} N

2

1.018 \times 10^{- 7} N

3

1.05 \times 10^{- 7} N

4

2.058 \times 10^{- 7} N

Explanation:

Here,
$r = 0.3 m m = 0.3 \times 10^{- 3} m, v = 1 m s^{- 1}$
$η = 18 \times 10^{- 5}$ poise
$= 18 \times 10^{- 6}$ deca poise
Viscous force, $F = 6 π η r v$
$= 6 \times \frac{22}{7} \times (18 \times 10^{- 6}) \times (0.3 \times 10^{- 3}) \times 1$
$= 1.018 \times 10^{- 7} N$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361405 Spherical balls of radius ' $R$ ' are falling in a viscous fluid of viscosity ' $η$ ' with a velocity ' $v$ '. The retarding viscous force acting on the spherical ball is

1 Inversely proportional to both radius '

R

' and velocity '

v

'

2 Directly proportional to both radius '

R

' and velocity '

v

'

3 Directly proportional to '

R

' but inversely proportional to '

v

'

4 Inversely proportional to '

R

' but Directly proportional to velocity '

v

'

Explanation:

From Stoke's law
Viscous force $F = 6 π η R v$
hence $F$ is directly proportional to radius & velocity.

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361406 When a ball is released from rest in a very long column of viscous liquid, its downward acceleration is ' $a$ ' (just after release). Then its acceleration when it has acquired two third of the maximum velocity:

1

\frac{a}{3}

2

\frac{2 a}{3}

3

\frac{a}{6}

4 None of these

Explanation:

When the ball is just released, the net force on ball is $m a = m g - F_{b}$
The terminal velocity ' $v_{t}$ ' of the ball is attained when net force on the ball is zero.
$m g = F_{b} + 6 π η r v_{t}$
When the ball acquires $\frac{2}{3} rd$ of its maximum velocity $v_{f}$
$\begin{aligned} m a^{'} = m g - F_{b} - 6 π η r (\frac{2 v_{t}}{3}) \\ m a^{'} = m a - \frac{2}{3} (m a) \\ a^{'} = \frac{a}{3} \end{aligned}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361407 Eight drops of water, each of radius $2 m m$ are falling through air at a terminal velocity of $8 c m s^{- 1}$ . If they coalesce to form a single drop, then the terminal velocity of combined drop will be

1

30 c m s^{- 1}

2

28 c m s^{- 1}

3

24 c m s^{- 1}

4

32 c m s^{- 1}

Explanation:

Let the radius of bigger drop is $R$ and that of smaller drop is $r$ then
$\frac{4}{3} π R^{3} = 8 \times \frac{4}{3} \times π r^{3}$
$or R = 2 r (1)$
Terminal velocity, $v \propto r^{2}$
$∴ \frac{v^{'}}{v} = \frac{R^{2}}{r^{2}} = {(\frac{2 r}{r})}^{2} = 4$
$\Rightarrow v^{'} = 4 v = 4 \times 8 = 32 c m s^{- 1}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361404 A rain drop of radius $0.3 m m$ has a terminal velocity in air $1 m s^{- 1} .$ The viscosity of air is $18 \times 10^{- 5}$ poise. Find the viscous force on the rain drops.

1

2.05 \times 10^{- 7} N

2

1.018 \times 10^{- 7} N

3

1.05 \times 10^{- 7} N

4

2.058 \times 10^{- 7} N

Explanation:

Here,
$r = 0.3 m m = 0.3 \times 10^{- 3} m, v = 1 m s^{- 1}$
$η = 18 \times 10^{- 5}$ poise
$= 18 \times 10^{- 6}$ deca poise
Viscous force, $F = 6 π η r v$
$= 6 \times \frac{22}{7} \times (18 \times 10^{- 6}) \times (0.3 \times 10^{- 3}) \times 1$
$= 1.018 \times 10^{- 7} N$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361405 Spherical balls of radius ' $R$ ' are falling in a viscous fluid of viscosity ' $η$ ' with a velocity ' $v$ '. The retarding viscous force acting on the spherical ball is

1 Inversely proportional to both radius '

R

' and velocity '

v

'

2 Directly proportional to both radius '

R

' and velocity '

v

'

3 Directly proportional to '

R

' but inversely proportional to '

v

'

4 Inversely proportional to '

R

' but Directly proportional to velocity '

v

'

Explanation:

From Stoke's law
Viscous force $F = 6 π η R v$
hence $F$ is directly proportional to radius & velocity.

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361406 When a ball is released from rest in a very long column of viscous liquid, its downward acceleration is ' $a$ ' (just after release). Then its acceleration when it has acquired two third of the maximum velocity:

1

\frac{a}{3}

2

\frac{2 a}{3}

3

\frac{a}{6}

4 None of these

Explanation:

When the ball is just released, the net force on ball is $m a = m g - F_{b}$
The terminal velocity ' $v_{t}$ ' of the ball is attained when net force on the ball is zero.
$m g = F_{b} + 6 π η r v_{t}$
When the ball acquires $\frac{2}{3} rd$ of its maximum velocity $v_{f}$
$\begin{aligned} m a^{'} = m g - F_{b} - 6 π η r (\frac{2 v_{t}}{3}) \\ m a^{'} = m a - \frac{2}{3} (m a) \\ a^{'} = \frac{a}{3} \end{aligned}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361407 Eight drops of water, each of radius $2 m m$ are falling through air at a terminal velocity of $8 c m s^{- 1}$ . If they coalesce to form a single drop, then the terminal velocity of combined drop will be

1

30 c m s^{- 1}

2

28 c m s^{- 1}

3

24 c m s^{- 1}

4

32 c m s^{- 1}

Explanation:

Let the radius of bigger drop is $R$ and that of smaller drop is $r$ then
$\frac{4}{3} π R^{3} = 8 \times \frac{4}{3} \times π r^{3}$
$or R = 2 r (1)$
Terminal velocity, $v \propto r^{2}$
$∴ \frac{v^{'}}{v} = \frac{R^{2}}{r^{2}} = {(\frac{2 r}{r})}^{2} = 4$
$\Rightarrow v^{'} = 4 v = 4 \times 8 = 32 c m s^{- 1}$

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