QBManager

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361360 Water flows in a streamlined manner through a capillary tube of radius $a$ , the pressure difference being $P$ and the rate of flow $Q$ . If the radius is reduced to $a / 2$ and the pressure increased to $2 P$ , the rate of flow becomes

1

Q

2

4 Q

3

\frac{Q}{8}

4

\frac{Q}{4}

Explanation:

$V = \frac{π Δ P r^{4}}{8 η l} ∴ V \propto Δ \Pr^{4}$
[ $η$ and $l$ are constants]
$\begin{aligned} ∴ \frac{V_{2}}{V_{1}} = (\frac{P_{2}}{P_{1}}) {(\frac{r_{2}}{r_{1}})}^{4} = 2 \times {(\frac{1}{2})}^{4} = \frac{1}{8} \\ ∴ V_{2} = \frac{Q}{8} \end{aligned}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361361 The formula for the resistance of a fluid is

1

R = \frac{8 η l}{π r^{2}}

2

R = \frac{π r^{4}}{8 η l}

3

R = \frac{8 η l}{π r^{4}}

4

R = \frac{8 η l}{π r^{3}}

Explanation:

The rate of liquid flow is
$Q = \frac{π Δ {Pr}^{4}}{8 η l} = \frac{Δ P}{(\frac{8 η l}{π r^{4}})}$
By comparing the above equation with ohm's law $i = \frac{Δ V}{R}$ then the resistance offered to the fluid is $R = \frac{8 η ℓ}{π r^{4}}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361362 A tube of length $l$ and radius $R$ carries a steady flow of fluid whose density is $ρ$ and viscosity $η$ . The fluid flow velocity depends on the distance $r$ from the axis of the tube as $v = v_{0} (1 - r^{2} / R^{2})$ . Find the mass of the fluid flowing across the section of the tube per unit time.

1

\frac{π ρ v_{0} R^{3}}{3}

2

2 π ρ v_{0} R^{2}

3

\frac{π ρ v_{0} R^{2}}{2}

4

\frac{π ρ v_{0} R^{3}}{4}

Explanation:

Consider the cross-section view of the tube.
supporting img

The small rate of flow of liquid across the circular of radius $r$ and width $d r$ is
$d Q = ρ (d A) v$
$\begin{aligned} d Q = ρ (2 π r) d r v = 2 π ρ v_{0} (r - \frac{r^{3}}{R^{2}}) d r \\ Q = 2 π ρ v_{0} [\int_{0}^{R} r d r - \frac{1}{R^{2}} \int_{0}^{R} r^{3} d r] \\ = 2 π ρ v_{0} [\frac{R^{2}}{2} - \frac{1}{R^{2}} \frac{R^{4}}{4}] = \frac{π ρ v_{0} R^{2}}{2} \end{aligned}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361363 Which factor controls the better flow rate of a liquid through the syringe?

1 The pressure exerted by the thumb

2 The length of the needle

3 The nature of the liquid

4 The radius of the syringe

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361360 Water flows in a streamlined manner through a capillary tube of radius $a$ , the pressure difference being $P$ and the rate of flow $Q$ . If the radius is reduced to $a / 2$ and the pressure increased to $2 P$ , the rate of flow becomes

1

Q

2

4 Q

3

\frac{Q}{8}

4

\frac{Q}{4}

Explanation:

$V = \frac{π Δ P r^{4}}{8 η l} ∴ V \propto Δ \Pr^{4}$
[ $η$ and $l$ are constants]
$\begin{aligned} ∴ \frac{V_{2}}{V_{1}} = (\frac{P_{2}}{P_{1}}) {(\frac{r_{2}}{r_{1}})}^{4} = 2 \times {(\frac{1}{2})}^{4} = \frac{1}{8} \\ ∴ V_{2} = \frac{Q}{8} \end{aligned}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361361 The formula for the resistance of a fluid is

1

R = \frac{8 η l}{π r^{2}}

2

R = \frac{π r^{4}}{8 η l}

3

R = \frac{8 η l}{π r^{4}}

4

R = \frac{8 η l}{π r^{3}}

Explanation:

The rate of liquid flow is
$Q = \frac{π Δ {Pr}^{4}}{8 η l} = \frac{Δ P}{(\frac{8 η l}{π r^{4}})}$
By comparing the above equation with ohm's law $i = \frac{Δ V}{R}$ then the resistance offered to the fluid is $R = \frac{8 η ℓ}{π r^{4}}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361362 A tube of length $l$ and radius $R$ carries a steady flow of fluid whose density is $ρ$ and viscosity $η$ . The fluid flow velocity depends on the distance $r$ from the axis of the tube as $v = v_{0} (1 - r^{2} / R^{2})$ . Find the mass of the fluid flowing across the section of the tube per unit time.

1

\frac{π ρ v_{0} R^{3}}{3}

2

2 π ρ v_{0} R^{2}

3

\frac{π ρ v_{0} R^{2}}{2}

4

\frac{π ρ v_{0} R^{3}}{4}

Explanation:

Consider the cross-section view of the tube.
supporting img

The small rate of flow of liquid across the circular of radius $r$ and width $d r$ is
$d Q = ρ (d A) v$
$\begin{aligned} d Q = ρ (2 π r) d r v = 2 π ρ v_{0} (r - \frac{r^{3}}{R^{2}}) d r \\ Q = 2 π ρ v_{0} [\int_{0}^{R} r d r - \frac{1}{R^{2}} \int_{0}^{R} r^{3} d r] \\ = 2 π ρ v_{0} [\frac{R^{2}}{2} - \frac{1}{R^{2}} \frac{R^{4}}{4}] = \frac{π ρ v_{0} R^{2}}{2} \end{aligned}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361363 Which factor controls the better flow rate of a liquid through the syringe?

1 The pressure exerted by the thumb

2 The length of the needle

3 The nature of the liquid

4 The radius of the syringe

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361360 Water flows in a streamlined manner through a capillary tube of radius $a$ , the pressure difference being $P$ and the rate of flow $Q$ . If the radius is reduced to $a / 2$ and the pressure increased to $2 P$ , the rate of flow becomes

1

Q

2

4 Q

3

\frac{Q}{8}

4

\frac{Q}{4}

Explanation:

$V = \frac{π Δ P r^{4}}{8 η l} ∴ V \propto Δ \Pr^{4}$
[ $η$ and $l$ are constants]
$\begin{aligned} ∴ \frac{V_{2}}{V_{1}} = (\frac{P_{2}}{P_{1}}) {(\frac{r_{2}}{r_{1}})}^{4} = 2 \times {(\frac{1}{2})}^{4} = \frac{1}{8} \\ ∴ V_{2} = \frac{Q}{8} \end{aligned}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361361 The formula for the resistance of a fluid is

1

R = \frac{8 η l}{π r^{2}}

2

R = \frac{π r^{4}}{8 η l}

3

R = \frac{8 η l}{π r^{4}}

4

R = \frac{8 η l}{π r^{3}}

Explanation:

The rate of liquid flow is
$Q = \frac{π Δ {Pr}^{4}}{8 η l} = \frac{Δ P}{(\frac{8 η l}{π r^{4}})}$
By comparing the above equation with ohm's law $i = \frac{Δ V}{R}$ then the resistance offered to the fluid is $R = \frac{8 η ℓ}{π r^{4}}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361362 A tube of length $l$ and radius $R$ carries a steady flow of fluid whose density is $ρ$ and viscosity $η$ . The fluid flow velocity depends on the distance $r$ from the axis of the tube as $v = v_{0} (1 - r^{2} / R^{2})$ . Find the mass of the fluid flowing across the section of the tube per unit time.

1

\frac{π ρ v_{0} R^{3}}{3}

2

2 π ρ v_{0} R^{2}

3

\frac{π ρ v_{0} R^{2}}{2}

4

\frac{π ρ v_{0} R^{3}}{4}

Explanation:

Consider the cross-section view of the tube.
supporting img

The small rate of flow of liquid across the circular of radius $r$ and width $d r$ is
$d Q = ρ (d A) v$
$\begin{aligned} d Q = ρ (2 π r) d r v = 2 π ρ v_{0} (r - \frac{r^{3}}{R^{2}}) d r \\ Q = 2 π ρ v_{0} [\int_{0}^{R} r d r - \frac{1}{R^{2}} \int_{0}^{R} r^{3} d r] \\ = 2 π ρ v_{0} [\frac{R^{2}}{2} - \frac{1}{R^{2}} \frac{R^{4}}{4}] = \frac{π ρ v_{0} R^{2}}{2} \end{aligned}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361363 Which factor controls the better flow rate of a liquid through the syringe?

1 The pressure exerted by the thumb

2 The length of the needle

3 The nature of the liquid

4 The radius of the syringe

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361360 Water flows in a streamlined manner through a capillary tube of radius $a$ , the pressure difference being $P$ and the rate of flow $Q$ . If the radius is reduced to $a / 2$ and the pressure increased to $2 P$ , the rate of flow becomes

1

Q

2

4 Q

3

\frac{Q}{8}

4

\frac{Q}{4}

Explanation:

$V = \frac{π Δ P r^{4}}{8 η l} ∴ V \propto Δ \Pr^{4}$
[ $η$ and $l$ are constants]
$\begin{aligned} ∴ \frac{V_{2}}{V_{1}} = (\frac{P_{2}}{P_{1}}) {(\frac{r_{2}}{r_{1}})}^{4} = 2 \times {(\frac{1}{2})}^{4} = \frac{1}{8} \\ ∴ V_{2} = \frac{Q}{8} \end{aligned}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361361 The formula for the resistance of a fluid is

1

R = \frac{8 η l}{π r^{2}}

2

R = \frac{π r^{4}}{8 η l}

3

R = \frac{8 η l}{π r^{4}}

4

R = \frac{8 η l}{π r^{3}}

Explanation:

The rate of liquid flow is
$Q = \frac{π Δ {Pr}^{4}}{8 η l} = \frac{Δ P}{(\frac{8 η l}{π r^{4}})}$
By comparing the above equation with ohm's law $i = \frac{Δ V}{R}$ then the resistance offered to the fluid is $R = \frac{8 η ℓ}{π r^{4}}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361362 A tube of length $l$ and radius $R$ carries a steady flow of fluid whose density is $ρ$ and viscosity $η$ . The fluid flow velocity depends on the distance $r$ from the axis of the tube as $v = v_{0} (1 - r^{2} / R^{2})$ . Find the mass of the fluid flowing across the section of the tube per unit time.

1

\frac{π ρ v_{0} R^{3}}{3}

2

2 π ρ v_{0} R^{2}

3

\frac{π ρ v_{0} R^{2}}{2}

4

\frac{π ρ v_{0} R^{3}}{4}

Explanation:

Consider the cross-section view of the tube.
supporting img

The small rate of flow of liquid across the circular of radius $r$ and width $d r$ is
$d Q = ρ (d A) v$
$\begin{aligned} d Q = ρ (2 π r) d r v = 2 π ρ v_{0} (r - \frac{r^{3}}{R^{2}}) d r \\ Q = 2 π ρ v_{0} [\int_{0}^{R} r d r - \frac{1}{R^{2}} \int_{0}^{R} r^{3} d r] \\ = 2 π ρ v_{0} [\frac{R^{2}}{2} - \frac{1}{R^{2}} \frac{R^{4}}{4}] = \frac{π ρ v_{0} R^{2}}{2} \end{aligned}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361363 Which factor controls the better flow rate of a liquid through the syringe?

1 The pressure exerted by the thumb

2 The length of the needle

3 The nature of the liquid

4 The radius of the syringe