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PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361395 The velocity of a ball of mass ' $m$ ' density ' $d_{1}$ ' when dropped in a container filled with glycerin of density ' $d_{2}$ ' becomes constant after sometime. The viscous force acting on the ball will be

1

m g (\frac{d_{1}}{d_{2}})

2

m g (1 - \frac{d_{2}}{d_{1}})

3

m g (\frac{d_{1} + d_{2}}{d_{1}})

4

m g (\frac{d_{1} + d_{2}}{d_{2}})

Explanation:

Viscous force $=$ Weight - Force of buoyancy
$\begin{matrix} \Rightarrow f_{v} = m g - F_{B} = m g - d_{2} v g \\ = m g [1 - \frac{d_{2} v g}{m g}] = m g [1 - \frac{d_{2}}{d_{1}}] \end{matrix}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361396 A sphere of mass $M$ and radius $R$ is falling in a viscous fluid. The terminal velocity attained by the object will be proportional to:

1

R

2

R^{2}

3

1 / R^{2}

4

1 / R

Explanation:

The terminal velocity of a sphere of radius $R$ moving in a liquid of coefficient of viscosity
$η is \Rightarrow v = \frac{2}{9} \frac{R^{2} (ρ - σ) g}{η}$
Where $ρ$ - density of the sphere
$σ$ -density of the liquid
$\Rightarrow v \propto R^{2}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361397 An air bubbles rises from te bottom of a lake of large depth. The rising speed of air bubble will

1 go on increasing till it reaches surface

2 go on decreasing till it reaches surface

3 increase in the beginning then will become constant

4 be constant all throughout

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361398 Eight equal drops of water are falling through air with a steady velocity of $10 c m s^{- 1}$ . If the drops combine to form a single drop big in size, then the terminal velocity of this big drop is

1

40 c m s^{- 1}

2

10 c m s^{- 1}

3

30 c m s^{- 1}

4

80 c m s^{- 1}

Explanation:

Let $r$ be radius of each small water droplet and $R$ be radius of a big drop.
Volume of big drop $= 8 \times$ Volume of each
$small droplets \frac{4}{3} π R^{3} = 8 \times \frac{4}{3} π r^{3} R = 2 r$
Let terminal velocity of small droplet be $v_{1}$ and of big drop be $v_{2}$ .
Terminal velocity, $v$ is
where the symbols have their usual meaning
$v μ r^{2}$
$\frac{v_{2}}{v_{1}} = \frac{r_{2}^{2}}{r_{1}^{2}} = {(\frac{2 r}{r})}^{2} = 4$
$\Rightarrow v_{2} = 4 v_{1}$
or $= 4 \times 10 c m s^{- 1} = 40 c m s^{- 1}$

KCET - 2011

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361395 The velocity of a ball of mass ' $m$ ' density ' $d_{1}$ ' when dropped in a container filled with glycerin of density ' $d_{2}$ ' becomes constant after sometime. The viscous force acting on the ball will be

1

m g (\frac{d_{1}}{d_{2}})

2

m g (1 - \frac{d_{2}}{d_{1}})

3

m g (\frac{d_{1} + d_{2}}{d_{1}})

4

m g (\frac{d_{1} + d_{2}}{d_{2}})

Explanation:

Viscous force $=$ Weight - Force of buoyancy
$\begin{matrix} \Rightarrow f_{v} = m g - F_{B} = m g - d_{2} v g \\ = m g [1 - \frac{d_{2} v g}{m g}] = m g [1 - \frac{d_{2}}{d_{1}}] \end{matrix}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361396 A sphere of mass $M$ and radius $R$ is falling in a viscous fluid. The terminal velocity attained by the object will be proportional to:

1

R

2

R^{2}

3

1 / R^{2}

4

1 / R

Explanation:

The terminal velocity of a sphere of radius $R$ moving in a liquid of coefficient of viscosity
$η is \Rightarrow v = \frac{2}{9} \frac{R^{2} (ρ - σ) g}{η}$
Where $ρ$ - density of the sphere
$σ$ -density of the liquid
$\Rightarrow v \propto R^{2}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361397 An air bubbles rises from te bottom of a lake of large depth. The rising speed of air bubble will

1 go on increasing till it reaches surface

2 go on decreasing till it reaches surface

3 increase in the beginning then will become constant

4 be constant all throughout

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361398 Eight equal drops of water are falling through air with a steady velocity of $10 c m s^{- 1}$ . If the drops combine to form a single drop big in size, then the terminal velocity of this big drop is

1

40 c m s^{- 1}

2

10 c m s^{- 1}

3

30 c m s^{- 1}

4

80 c m s^{- 1}

Explanation:

Let $r$ be radius of each small water droplet and $R$ be radius of a big drop.
Volume of big drop $= 8 \times$ Volume of each
$small droplets \frac{4}{3} π R^{3} = 8 \times \frac{4}{3} π r^{3} R = 2 r$
Let terminal velocity of small droplet be $v_{1}$ and of big drop be $v_{2}$ .
Terminal velocity, $v$ is
where the symbols have their usual meaning
$v μ r^{2}$
$\frac{v_{2}}{v_{1}} = \frac{r_{2}^{2}}{r_{1}^{2}} = {(\frac{2 r}{r})}^{2} = 4$
$\Rightarrow v_{2} = 4 v_{1}$
or $= 4 \times 10 c m s^{- 1} = 40 c m s^{- 1}$

KCET - 2011

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361395 The velocity of a ball of mass ' $m$ ' density ' $d_{1}$ ' when dropped in a container filled with glycerin of density ' $d_{2}$ ' becomes constant after sometime. The viscous force acting on the ball will be

1

m g (\frac{d_{1}}{d_{2}})

2

m g (1 - \frac{d_{2}}{d_{1}})

3

m g (\frac{d_{1} + d_{2}}{d_{1}})

4

m g (\frac{d_{1} + d_{2}}{d_{2}})

Explanation:

Viscous force $=$ Weight - Force of buoyancy
$\begin{matrix} \Rightarrow f_{v} = m g - F_{B} = m g - d_{2} v g \\ = m g [1 - \frac{d_{2} v g}{m g}] = m g [1 - \frac{d_{2}}{d_{1}}] \end{matrix}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361396 A sphere of mass $M$ and radius $R$ is falling in a viscous fluid. The terminal velocity attained by the object will be proportional to:

1

R

2

R^{2}

3

1 / R^{2}

4

1 / R

Explanation:

The terminal velocity of a sphere of radius $R$ moving in a liquid of coefficient of viscosity
$η is \Rightarrow v = \frac{2}{9} \frac{R^{2} (ρ - σ) g}{η}$
Where $ρ$ - density of the sphere
$σ$ -density of the liquid
$\Rightarrow v \propto R^{2}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361397 An air bubbles rises from te bottom of a lake of large depth. The rising speed of air bubble will

1 go on increasing till it reaches surface

2 go on decreasing till it reaches surface

3 increase in the beginning then will become constant

4 be constant all throughout

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361398 Eight equal drops of water are falling through air with a steady velocity of $10 c m s^{- 1}$ . If the drops combine to form a single drop big in size, then the terminal velocity of this big drop is

1

40 c m s^{- 1}

2

10 c m s^{- 1}

3

30 c m s^{- 1}

4

80 c m s^{- 1}

Explanation:

Let $r$ be radius of each small water droplet and $R$ be radius of a big drop.
Volume of big drop $= 8 \times$ Volume of each
$small droplets \frac{4}{3} π R^{3} = 8 \times \frac{4}{3} π r^{3} R = 2 r$
Let terminal velocity of small droplet be $v_{1}$ and of big drop be $v_{2}$ .
Terminal velocity, $v$ is
where the symbols have their usual meaning
$v μ r^{2}$
$\frac{v_{2}}{v_{1}} = \frac{r_{2}^{2}}{r_{1}^{2}} = {(\frac{2 r}{r})}^{2} = 4$
$\Rightarrow v_{2} = 4 v_{1}$
or $= 4 \times 10 c m s^{- 1} = 40 c m s^{- 1}$

KCET - 2011

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361395 The velocity of a ball of mass ' $m$ ' density ' $d_{1}$ ' when dropped in a container filled with glycerin of density ' $d_{2}$ ' becomes constant after sometime. The viscous force acting on the ball will be

1

m g (\frac{d_{1}}{d_{2}})

2

m g (1 - \frac{d_{2}}{d_{1}})

3

m g (\frac{d_{1} + d_{2}}{d_{1}})

4

m g (\frac{d_{1} + d_{2}}{d_{2}})

Explanation:

Viscous force $=$ Weight - Force of buoyancy
$\begin{matrix} \Rightarrow f_{v} = m g - F_{B} = m g - d_{2} v g \\ = m g [1 - \frac{d_{2} v g}{m g}] = m g [1 - \frac{d_{2}}{d_{1}}] \end{matrix}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361396 A sphere of mass $M$ and radius $R$ is falling in a viscous fluid. The terminal velocity attained by the object will be proportional to:

1

R

2

R^{2}

3

1 / R^{2}

4

1 / R

Explanation:

The terminal velocity of a sphere of radius $R$ moving in a liquid of coefficient of viscosity
$η is \Rightarrow v = \frac{2}{9} \frac{R^{2} (ρ - σ) g}{η}$
Where $ρ$ - density of the sphere
$σ$ -density of the liquid
$\Rightarrow v \propto R^{2}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361397 An air bubbles rises from te bottom of a lake of large depth. The rising speed of air bubble will

1 go on increasing till it reaches surface

2 go on decreasing till it reaches surface

3 increase in the beginning then will become constant

4 be constant all throughout

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361398 Eight equal drops of water are falling through air with a steady velocity of $10 c m s^{- 1}$ . If the drops combine to form a single drop big in size, then the terminal velocity of this big drop is

1

40 c m s^{- 1}

2

10 c m s^{- 1}

3

30 c m s^{- 1}

4

80 c m s^{- 1}

Explanation:

Let $r$ be radius of each small water droplet and $R$ be radius of a big drop.
Volume of big drop $= 8 \times$ Volume of each
$small droplets \frac{4}{3} π R^{3} = 8 \times \frac{4}{3} π r^{3} R = 2 r$
Let terminal velocity of small droplet be $v_{1}$ and of big drop be $v_{2}$ .
Terminal velocity, $v$ is
where the symbols have their usual meaning
$v μ r^{2}$
$\frac{v_{2}}{v_{1}} = \frac{r_{2}^{2}}{r_{1}^{2}} = {(\frac{2 r}{r})}^{2} = 4$
$\Rightarrow v_{2} = 4 v_{1}$
or $= 4 \times 10 c m s^{- 1} = 40 c m s^{- 1}$

KCET - 2011