QBManager

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361334 A long cylinder of radius $R_{1}$ is displaced along its axis with a constant velocity $v_{0}$ inside a stationary co-axial cylinder of radius $R_{2}$ as shown in the figure. The space between the cylinders is filled with viscous liquid. Find the velocity of the liquid as a function of the distance $r$ from the axis of the cylinders.
supporting img

1

+ v_{0} \ln (\frac{R_{2}}{r})

2

\frac{v_{0}}{\ln (\frac{R_{2}}{R_{1}})}

3

v_{0} \ln (\frac{R_{2}}{R_{1}})

4

v_{0} \frac{\ln (R_{2} / r)}{\ln (R_{2} / R_{1})}

Explanation:

Let $v$ is the velocity of a spherical shell or radius $r$
The viscous force between two adjacent layers having radius $r$ is
supporting img

where $L$ is the length of the cylinder. $d v$ is the relative velocity between two layers which are separated by $d r$ .
As we move towards right $v$ decreases and $r$ increases.
$d v = (-) v e & d r = (+) v e$
$f = - 2 π r L η \frac{d v}{d r}$
$\int_{R_{1}}^{r} \frac{d r}{r} = \frac{- 2 π L η}{f} \int_{v_{0}}^{v} d v$
$\ln \frac{r}{R_{1}} = \frac{2 π L η (v_{0} - v)}{f} (1)$
$\int_{R_{1}}^{R_{2}} \frac{d r}{r} = \frac{- 2 π L η}{f} \int_{v_{0}}^{0} d v$
$\ln \frac{R_{2}}{R_{1}} = \frac{2 π L η v_{0}}{f} (2)$
From the above two equations we get
$\frac{\ln (r / R_{1})}{\ln (R_{2} / R_{1})} = (\frac{v_{0} - v}{v_{0}})$
$v = v_{0} [1 - \frac{\ln (r / R_{1})}{\ln (R_{2} / R_{1})}] = \frac{v_{0} \ln (R_{2} / r)}{\ln (R_{2} / R_{1})}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361335 Figure shows a long solid cylinder of radius $R$ inside a long hollow tube of inner radius $R^{'}$ . A viscous liquid of coefficient of viscosity $η$ is filled in the gap between the cylinder and the hollow tube. If the solid cylinder is moved with velocity $v$ parallel to its length inside the fixed hollow tube, find the force required per unit length to be applied on the solid cylinder, assuming uniform velocity gradient in the liquid.
supporting img

1

η (2 π R) (\frac{v}{R^{'} - R})

2

η (2 π R^{'}) (\frac{v}{R^{'} - R})

3

η (4 π R) (\frac{v}{R^{'} + R})

4

η (4 π R^{'}) (\frac{v}{R^{'} + R})

Explanation:

The viscous force on the cylinder is
$\begin{aligned} F = - η A \frac{d v}{d x} = - (2 π R h η) \frac{v - 0}{R - R^{'}} \\ \Rightarrow \frac{F}{h} = η (2 π R) \frac{v}{R^{'} - R} \end{aligned}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361336 A layer of glycerin of thinkness $3 m m$ is present between a large surface and a surface of area of $0.2 m^{2}$ , with what force the small surface is to be pulled, so that it can move with a velocity of $5 m / s$ ? $(η = 0.09 k g m^{- 1} s^{- 1})$ .

1

10 N

2

30 N

3

40 N

4

50 N

Explanation:

$η = 0.09 k g m^{- 1} s^{- 1}, d v = 5 m / s,$
$d x = 3 \times 10^{- 3} m, A = 0.2 m^{2}$
$F = η \frac{A d v}{d x} = \frac{0.09 \times 0.2 \times 5}{3 \times 10^{- 3}} = 30 N$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361337 The velocity of water in a river is $18 k m / h r$ near the surface. If the river is $5 m$ deep, find the shearing stress between the horizontal layers of water. The co-efficient of viscosity of water $= 10^{- 2}$ poise.

1

10^{- 1} N / m^{2}

2

10^{- 2} N / m^{2}

3

10^{- 3} N / m^{2}

4

10^{- 4} N / m^{2}

Explanation:

$η = 10^{- 2} p o i s e = 10^{- 3} \frac{N}{m^{2}} s$
$v = 18 k m / h = \frac{18000}{3600} = 5 m / s$
$l = 5 m$
Strain rate $= \frac{v}{l}$
Coefficient of viscosity, $η = \frac{shearing stress}{strain rate}$
$∴$ Shearing strees $= η \times$ strain rate
$= 10^{- 3} \times \frac{5}{5} = 10^{- 3} N m^{- 2}$

JEE - 2014

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361334 A long cylinder of radius $R_{1}$ is displaced along its axis with a constant velocity $v_{0}$ inside a stationary co-axial cylinder of radius $R_{2}$ as shown in the figure. The space between the cylinders is filled with viscous liquid. Find the velocity of the liquid as a function of the distance $r$ from the axis of the cylinders.
supporting img

1

+ v_{0} \ln (\frac{R_{2}}{r})

2

\frac{v_{0}}{\ln (\frac{R_{2}}{R_{1}})}

3

v_{0} \ln (\frac{R_{2}}{R_{1}})

4

v_{0} \frac{\ln (R_{2} / r)}{\ln (R_{2} / R_{1})}

Explanation:

Let $v$ is the velocity of a spherical shell or radius $r$
The viscous force between two adjacent layers having radius $r$ is
supporting img

where $L$ is the length of the cylinder. $d v$ is the relative velocity between two layers which are separated by $d r$ .
As we move towards right $v$ decreases and $r$ increases.
$d v = (-) v e & d r = (+) v e$
$f = - 2 π r L η \frac{d v}{d r}$
$\int_{R_{1}}^{r} \frac{d r}{r} = \frac{- 2 π L η}{f} \int_{v_{0}}^{v} d v$
$\ln \frac{r}{R_{1}} = \frac{2 π L η (v_{0} - v)}{f} (1)$
$\int_{R_{1}}^{R_{2}} \frac{d r}{r} = \frac{- 2 π L η}{f} \int_{v_{0}}^{0} d v$
$\ln \frac{R_{2}}{R_{1}} = \frac{2 π L η v_{0}}{f} (2)$
From the above two equations we get
$\frac{\ln (r / R_{1})}{\ln (R_{2} / R_{1})} = (\frac{v_{0} - v}{v_{0}})$
$v = v_{0} [1 - \frac{\ln (r / R_{1})}{\ln (R_{2} / R_{1})}] = \frac{v_{0} \ln (R_{2} / r)}{\ln (R_{2} / R_{1})}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361335 Figure shows a long solid cylinder of radius $R$ inside a long hollow tube of inner radius $R^{'}$ . A viscous liquid of coefficient of viscosity $η$ is filled in the gap between the cylinder and the hollow tube. If the solid cylinder is moved with velocity $v$ parallel to its length inside the fixed hollow tube, find the force required per unit length to be applied on the solid cylinder, assuming uniform velocity gradient in the liquid.
supporting img

1

η (2 π R) (\frac{v}{R^{'} - R})

2

η (2 π R^{'}) (\frac{v}{R^{'} - R})

3

η (4 π R) (\frac{v}{R^{'} + R})

4

η (4 π R^{'}) (\frac{v}{R^{'} + R})

Explanation:

The viscous force on the cylinder is
$\begin{aligned} F = - η A \frac{d v}{d x} = - (2 π R h η) \frac{v - 0}{R - R^{'}} \\ \Rightarrow \frac{F}{h} = η (2 π R) \frac{v}{R^{'} - R} \end{aligned}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361336 A layer of glycerin of thinkness $3 m m$ is present between a large surface and a surface of area of $0.2 m^{2}$ , with what force the small surface is to be pulled, so that it can move with a velocity of $5 m / s$ ? $(η = 0.09 k g m^{- 1} s^{- 1})$ .

1

10 N

2

30 N

3

40 N

4

50 N

Explanation:

$η = 0.09 k g m^{- 1} s^{- 1}, d v = 5 m / s,$
$d x = 3 \times 10^{- 3} m, A = 0.2 m^{2}$
$F = η \frac{A d v}{d x} = \frac{0.09 \times 0.2 \times 5}{3 \times 10^{- 3}} = 30 N$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361337 The velocity of water in a river is $18 k m / h r$ near the surface. If the river is $5 m$ deep, find the shearing stress between the horizontal layers of water. The co-efficient of viscosity of water $= 10^{- 2}$ poise.

1

10^{- 1} N / m^{2}

2

10^{- 2} N / m^{2}

3

10^{- 3} N / m^{2}

4

10^{- 4} N / m^{2}

Explanation:

$η = 10^{- 2} p o i s e = 10^{- 3} \frac{N}{m^{2}} s$
$v = 18 k m / h = \frac{18000}{3600} = 5 m / s$
$l = 5 m$
Strain rate $= \frac{v}{l}$
Coefficient of viscosity, $η = \frac{shearing stress}{strain rate}$
$∴$ Shearing strees $= η \times$ strain rate
$= 10^{- 3} \times \frac{5}{5} = 10^{- 3} N m^{- 2}$

JEE - 2014

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361334 A long cylinder of radius $R_{1}$ is displaced along its axis with a constant velocity $v_{0}$ inside a stationary co-axial cylinder of radius $R_{2}$ as shown in the figure. The space between the cylinders is filled with viscous liquid. Find the velocity of the liquid as a function of the distance $r$ from the axis of the cylinders.
supporting img

1

+ v_{0} \ln (\frac{R_{2}}{r})

2

\frac{v_{0}}{\ln (\frac{R_{2}}{R_{1}})}

3

v_{0} \ln (\frac{R_{2}}{R_{1}})

4

v_{0} \frac{\ln (R_{2} / r)}{\ln (R_{2} / R_{1})}

Explanation:

Let $v$ is the velocity of a spherical shell or radius $r$
The viscous force between two adjacent layers having radius $r$ is
supporting img

where $L$ is the length of the cylinder. $d v$ is the relative velocity between two layers which are separated by $d r$ .
As we move towards right $v$ decreases and $r$ increases.
$d v = (-) v e & d r = (+) v e$
$f = - 2 π r L η \frac{d v}{d r}$
$\int_{R_{1}}^{r} \frac{d r}{r} = \frac{- 2 π L η}{f} \int_{v_{0}}^{v} d v$
$\ln \frac{r}{R_{1}} = \frac{2 π L η (v_{0} - v)}{f} (1)$
$\int_{R_{1}}^{R_{2}} \frac{d r}{r} = \frac{- 2 π L η}{f} \int_{v_{0}}^{0} d v$
$\ln \frac{R_{2}}{R_{1}} = \frac{2 π L η v_{0}}{f} (2)$
From the above two equations we get
$\frac{\ln (r / R_{1})}{\ln (R_{2} / R_{1})} = (\frac{v_{0} - v}{v_{0}})$
$v = v_{0} [1 - \frac{\ln (r / R_{1})}{\ln (R_{2} / R_{1})}] = \frac{v_{0} \ln (R_{2} / r)}{\ln (R_{2} / R_{1})}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361335 Figure shows a long solid cylinder of radius $R$ inside a long hollow tube of inner radius $R^{'}$ . A viscous liquid of coefficient of viscosity $η$ is filled in the gap between the cylinder and the hollow tube. If the solid cylinder is moved with velocity $v$ parallel to its length inside the fixed hollow tube, find the force required per unit length to be applied on the solid cylinder, assuming uniform velocity gradient in the liquid.
supporting img

1

η (2 π R) (\frac{v}{R^{'} - R})

2

η (2 π R^{'}) (\frac{v}{R^{'} - R})

3

η (4 π R) (\frac{v}{R^{'} + R})

4

η (4 π R^{'}) (\frac{v}{R^{'} + R})

Explanation:

The viscous force on the cylinder is
$\begin{aligned} F = - η A \frac{d v}{d x} = - (2 π R h η) \frac{v - 0}{R - R^{'}} \\ \Rightarrow \frac{F}{h} = η (2 π R) \frac{v}{R^{'} - R} \end{aligned}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361336 A layer of glycerin of thinkness $3 m m$ is present between a large surface and a surface of area of $0.2 m^{2}$ , with what force the small surface is to be pulled, so that it can move with a velocity of $5 m / s$ ? $(η = 0.09 k g m^{- 1} s^{- 1})$ .

1

10 N

2

30 N

3

40 N

4

50 N

Explanation:

$η = 0.09 k g m^{- 1} s^{- 1}, d v = 5 m / s,$
$d x = 3 \times 10^{- 3} m, A = 0.2 m^{2}$
$F = η \frac{A d v}{d x} = \frac{0.09 \times 0.2 \times 5}{3 \times 10^{- 3}} = 30 N$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361337 The velocity of water in a river is $18 k m / h r$ near the surface. If the river is $5 m$ deep, find the shearing stress between the horizontal layers of water. The co-efficient of viscosity of water $= 10^{- 2}$ poise.

1

10^{- 1} N / m^{2}

2

10^{- 2} N / m^{2}

3

10^{- 3} N / m^{2}

4

10^{- 4} N / m^{2}

Explanation:

$η = 10^{- 2} p o i s e = 10^{- 3} \frac{N}{m^{2}} s$
$v = 18 k m / h = \frac{18000}{3600} = 5 m / s$
$l = 5 m$
Strain rate $= \frac{v}{l}$
Coefficient of viscosity, $η = \frac{shearing stress}{strain rate}$
$∴$ Shearing strees $= η \times$ strain rate
$= 10^{- 3} \times \frac{5}{5} = 10^{- 3} N m^{- 2}$

JEE - 2014

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361334 A long cylinder of radius $R_{1}$ is displaced along its axis with a constant velocity $v_{0}$ inside a stationary co-axial cylinder of radius $R_{2}$ as shown in the figure. The space between the cylinders is filled with viscous liquid. Find the velocity of the liquid as a function of the distance $r$ from the axis of the cylinders.
supporting img

1

+ v_{0} \ln (\frac{R_{2}}{r})

2

\frac{v_{0}}{\ln (\frac{R_{2}}{R_{1}})}

3

v_{0} \ln (\frac{R_{2}}{R_{1}})

4

v_{0} \frac{\ln (R_{2} / r)}{\ln (R_{2} / R_{1})}

Explanation:

Let $v$ is the velocity of a spherical shell or radius $r$
The viscous force between two adjacent layers having radius $r$ is
supporting img

where $L$ is the length of the cylinder. $d v$ is the relative velocity between two layers which are separated by $d r$ .
As we move towards right $v$ decreases and $r$ increases.
$d v = (-) v e & d r = (+) v e$
$f = - 2 π r L η \frac{d v}{d r}$
$\int_{R_{1}}^{r} \frac{d r}{r} = \frac{- 2 π L η}{f} \int_{v_{0}}^{v} d v$
$\ln \frac{r}{R_{1}} = \frac{2 π L η (v_{0} - v)}{f} (1)$
$\int_{R_{1}}^{R_{2}} \frac{d r}{r} = \frac{- 2 π L η}{f} \int_{v_{0}}^{0} d v$
$\ln \frac{R_{2}}{R_{1}} = \frac{2 π L η v_{0}}{f} (2)$
From the above two equations we get
$\frac{\ln (r / R_{1})}{\ln (R_{2} / R_{1})} = (\frac{v_{0} - v}{v_{0}})$
$v = v_{0} [1 - \frac{\ln (r / R_{1})}{\ln (R_{2} / R_{1})}] = \frac{v_{0} \ln (R_{2} / r)}{\ln (R_{2} / R_{1})}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361335 Figure shows a long solid cylinder of radius $R$ inside a long hollow tube of inner radius $R^{'}$ . A viscous liquid of coefficient of viscosity $η$ is filled in the gap between the cylinder and the hollow tube. If the solid cylinder is moved with velocity $v$ parallel to its length inside the fixed hollow tube, find the force required per unit length to be applied on the solid cylinder, assuming uniform velocity gradient in the liquid.
supporting img

1

η (2 π R) (\frac{v}{R^{'} - R})

2

η (2 π R^{'}) (\frac{v}{R^{'} - R})

3

η (4 π R) (\frac{v}{R^{'} + R})

4

η (4 π R^{'}) (\frac{v}{R^{'} + R})

Explanation:

The viscous force on the cylinder is
$\begin{aligned} F = - η A \frac{d v}{d x} = - (2 π R h η) \frac{v - 0}{R - R^{'}} \\ \Rightarrow \frac{F}{h} = η (2 π R) \frac{v}{R^{'} - R} \end{aligned}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361336 A layer of glycerin of thinkness $3 m m$ is present between a large surface and a surface of area of $0.2 m^{2}$ , with what force the small surface is to be pulled, so that it can move with a velocity of $5 m / s$ ? $(η = 0.09 k g m^{- 1} s^{- 1})$ .

1

10 N

2

30 N

3

40 N

4

50 N

Explanation:

$η = 0.09 k g m^{- 1} s^{- 1}, d v = 5 m / s,$
$d x = 3 \times 10^{- 3} m, A = 0.2 m^{2}$
$F = η \frac{A d v}{d x} = \frac{0.09 \times 0.2 \times 5}{3 \times 10^{- 3}} = 30 N$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361337 The velocity of water in a river is $18 k m / h r$ near the surface. If the river is $5 m$ deep, find the shearing stress between the horizontal layers of water. The co-efficient of viscosity of water $= 10^{- 2}$ poise.

1

10^{- 1} N / m^{2}

2

10^{- 2} N / m^{2}

3

10^{- 3} N / m^{2}

4

10^{- 4} N / m^{2}

Explanation:

$η = 10^{- 2} p o i s e = 10^{- 3} \frac{N}{m^{2}} s$
$v = 18 k m / h = \frac{18000}{3600} = 5 m / s$
$l = 5 m$
Strain rate $= \frac{v}{l}$
Coefficient of viscosity, $η = \frac{shearing stress}{strain rate}$
$∴$ Shearing strees $= η \times$ strain rate
$= 10^{- 3} \times \frac{5}{5} = 10^{- 3} N m^{- 2}$

JEE - 2014

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation: