QBManager

Mechanical Properties of Fluids

142723 A body of density $d_{1}$ is counter poised by $M g$ of weights of density $d_{2}$ in air of density $d$ . Then the true mass of the body is

1

M

2

M (\frac{d_{2} - d}{d_{2}})

3

M (\frac{d_{1} - d}{d_{1}})

4

\frac{{Md}_{1} (d_{2} - d)}{d_{2} (d_{1} - d)}

Explanation:

D Apparent Weight (A.W) = True Weight Weight of liquid displaced
$AW = Mg - \frac{M}{d_{2}} \cdot dg$
$AW = M_{o} g - \frac{M_{o}}{d_{1}} \cdot dg$
From equation (i) and (ii), we get -
$M g - \frac{M}{d_{2}} d g = M_{o} g - \frac{M_{o}}{d_{1}} d g$
$\Rightarrow M_{o} (1 - \frac{d}{d_{1}}) = M (1 - \frac{d}{d_{2}})$
$\Rightarrow M_{o} = \frac{M (1 - \frac{d}{d_{2}})}{1 - \frac{d}{d_{1}}}$
$\Rightarrow M_{o} = \frac{M_{1} (d_{2} - d)}{d_{2} (d_{1} - d)}$

AP EAMCET-24.09.2020

Mechanical Properties of Fluids

142725 A metal plate of area $100 {cm}^{2}$ is lying on a liquid layer of thickness $2 mm$ . If the coefficient of viscosity of the liquid is $2 N$ . $s^{- 2} m^{- 2}$ , then find the minimum horizontal force required to move the plate with a speed of $1 cm . s^{- 1}$

1

0.1 N

2

1 N

3

0.5 N

4

0.25 N

Explanation:

A Given that,
Area (A) $= 100 {cm}^{2} = 100 \times 10^{- 4} m^{2}$
Thickness (dy) $= 2 mm = 2 \times 10^{- 3} m$
Coefficient of viscosity $= 2$ N.s.m $^{- 2}$
$Velocity = 1 cm s^{- 1} = 1 \times 10^{- 2} {ms}^{- 1}$
$F = η A (\frac{dv}{dy})$
$∴ F = \frac{2 \times 100 \times 10^{- 4} \times 1 \times 10^{- 2}}{2 \times 10^{- 3}} = 10^{- 1}$
$F = \frac{1}{10} = 0.1 N$

AP EAMCET (Medical)-07.10.2020

Mechanical Properties of Fluids

142727 Find the Young's modulus of the wire whose stress-strain curve is as shown in the following figure:

1

8 \times 10^{11} N . m^{- 2}

2

24 \times 10^{11} N \cdot m^{- 2}

3

10 \times 10^{11} N \cdot m^{- 2}

4

2 \times 10^{11} N \cdot m^{- 2}

Explanation:

D From above graph it is clear that stress $=$ $8 \times 10^{7} N / m^{2}$ and strain $= 4 \times 10^{- 4}$
We know that,
$Young's Modulus (Y) = \frac{Stress}{Strain}$
$Y = \frac{8 \times 10^{7}}{4 \times 10^{- 4}} = 2 \times 10^{11} N / m^{2}$

AP EAMCET-24.08.2021

Mechanical Properties of Fluids

142728 An air bubble of radius $1.0 cm$ rises with a constant speed of $3.5 mm s^{- 1}$ through a liquid of density $1.75 \times 10^{3} {kgm}^{- 3}$ . Neglecting the density of air, the coefficient of viscosity of the liquid is ${k g m}^{- 1} s^{- 1}$

1 54.5

2 109

3 163.5

4 218

Explanation:

B Given that,
Radius of air bubble $= 1.0 cm = 1 \times 10^{- 2} m$
Speed $(v) = 3.5 mm / s = 3.5 \times 10^{- 3} m / s$
Density $(ρ) = 1.75 \times 10^{3} kg / m^{3}$
Coefficient of viscosity $(η) =$ ?
We know that,
Terminal velocity $= \frac{2 r^{2} (ρ) g}{9 η}$
$3.5 \times 10^{- 3} = \frac{2 \times 10^{- 4} \times 1.75 \times 10^{3} \times 9.8}{9 \times η}$
$η = \frac{2 \times 10^{- 4} \times 1.75 \times 10^{3} \times 9.8}{9 \times 3.5 \times 10^{- 3}}$
$η = 108.88 ≃ 109 {kgm}^{- 1} s^{- 1}$

AP EAMCET-03.09.2021

Mechanical Properties of Fluids

142729 A flat plate of area $10 {cm}^{2}$ is separated and a large plate by a layer of Glycerine $1 mm$ thick. If the coefficient of viscosity of Glycerine is $20$ poise. The force required to keep the plate moving the velocity of $1 cm . s^{- 1}$ is

1 80 dyne

2 200 dyne

3 800 dyne

4 2000 dyne

Explanation:

D Given that,
$Area (A) = 10 {cm}^{2}$
Coefficient of viscosity of Glycerine $= 20$ poise Velocity $= 1 cm / s$
We know that,
Force required $= η A \frac{dv}{dy}$
Now $\frac{dv}{dy} = \frac{1 {cms}^{- 1}}{1 \times 10^{- 1} cm} = 10 s^{- 1}$
So, force to keep plate moving $= 10 \times 10 \times 20$ dyne
$= 2000 dyne$

AP EAMCET-06.09.2021

Mechanical Properties of Fluids

142723 A body of density $d_{1}$ is counter poised by $M g$ of weights of density $d_{2}$ in air of density $d$ . Then the true mass of the body is

1

M

2

M (\frac{d_{2} - d}{d_{2}})

3

M (\frac{d_{1} - d}{d_{1}})

4

\frac{{Md}_{1} (d_{2} - d)}{d_{2} (d_{1} - d)}

Explanation:

D Apparent Weight (A.W) = True Weight Weight of liquid displaced
$AW = Mg - \frac{M}{d_{2}} \cdot dg$
$AW = M_{o} g - \frac{M_{o}}{d_{1}} \cdot dg$
From equation (i) and (ii), we get -
$M g - \frac{M}{d_{2}} d g = M_{o} g - \frac{M_{o}}{d_{1}} d g$
$\Rightarrow M_{o} (1 - \frac{d}{d_{1}}) = M (1 - \frac{d}{d_{2}})$
$\Rightarrow M_{o} = \frac{M (1 - \frac{d}{d_{2}})}{1 - \frac{d}{d_{1}}}$
$\Rightarrow M_{o} = \frac{M_{1} (d_{2} - d)}{d_{2} (d_{1} - d)}$

AP EAMCET-24.09.2020

Mechanical Properties of Fluids

142725 A metal plate of area $100 {cm}^{2}$ is lying on a liquid layer of thickness $2 mm$ . If the coefficient of viscosity of the liquid is $2 N$ . $s^{- 2} m^{- 2}$ , then find the minimum horizontal force required to move the plate with a speed of $1 cm . s^{- 1}$

1

0.1 N

2

1 N

3

0.5 N

4

0.25 N

Explanation:

A Given that,
Area (A) $= 100 {cm}^{2} = 100 \times 10^{- 4} m^{2}$
Thickness (dy) $= 2 mm = 2 \times 10^{- 3} m$
Coefficient of viscosity $= 2$ N.s.m $^{- 2}$
$Velocity = 1 cm s^{- 1} = 1 \times 10^{- 2} {ms}^{- 1}$
$F = η A (\frac{dv}{dy})$
$∴ F = \frac{2 \times 100 \times 10^{- 4} \times 1 \times 10^{- 2}}{2 \times 10^{- 3}} = 10^{- 1}$
$F = \frac{1}{10} = 0.1 N$

AP EAMCET (Medical)-07.10.2020

Mechanical Properties of Fluids

142727 Find the Young's modulus of the wire whose stress-strain curve is as shown in the following figure:

1

8 \times 10^{11} N . m^{- 2}

2

24 \times 10^{11} N \cdot m^{- 2}

3

10 \times 10^{11} N \cdot m^{- 2}

4

2 \times 10^{11} N \cdot m^{- 2}

Explanation:

D From above graph it is clear that stress $=$ $8 \times 10^{7} N / m^{2}$ and strain $= 4 \times 10^{- 4}$
We know that,
$Young's Modulus (Y) = \frac{Stress}{Strain}$
$Y = \frac{8 \times 10^{7}}{4 \times 10^{- 4}} = 2 \times 10^{11} N / m^{2}$

AP EAMCET-24.08.2021

Mechanical Properties of Fluids

142728 An air bubble of radius $1.0 cm$ rises with a constant speed of $3.5 mm s^{- 1}$ through a liquid of density $1.75 \times 10^{3} {kgm}^{- 3}$ . Neglecting the density of air, the coefficient of viscosity of the liquid is ${k g m}^{- 1} s^{- 1}$

1 54.5

2 109

3 163.5

4 218

Explanation:

B Given that,
Radius of air bubble $= 1.0 cm = 1 \times 10^{- 2} m$
Speed $(v) = 3.5 mm / s = 3.5 \times 10^{- 3} m / s$
Density $(ρ) = 1.75 \times 10^{3} kg / m^{3}$
Coefficient of viscosity $(η) =$ ?
We know that,
Terminal velocity $= \frac{2 r^{2} (ρ) g}{9 η}$
$3.5 \times 10^{- 3} = \frac{2 \times 10^{- 4} \times 1.75 \times 10^{3} \times 9.8}{9 \times η}$
$η = \frac{2 \times 10^{- 4} \times 1.75 \times 10^{3} \times 9.8}{9 \times 3.5 \times 10^{- 3}}$
$η = 108.88 ≃ 109 {kgm}^{- 1} s^{- 1}$

AP EAMCET-03.09.2021

Mechanical Properties of Fluids

142729 A flat plate of area $10 {cm}^{2}$ is separated and a large plate by a layer of Glycerine $1 mm$ thick. If the coefficient of viscosity of Glycerine is $20$ poise. The force required to keep the plate moving the velocity of $1 cm . s^{- 1}$ is

1 80 dyne

2 200 dyne

3 800 dyne

4 2000 dyne

Explanation:

D Given that,
$Area (A) = 10 {cm}^{2}$
Coefficient of viscosity of Glycerine $= 20$ poise Velocity $= 1 cm / s$
We know that,
Force required $= η A \frac{dv}{dy}$
Now $\frac{dv}{dy} = \frac{1 {cms}^{- 1}}{1 \times 10^{- 1} cm} = 10 s^{- 1}$
So, force to keep plate moving $= 10 \times 10 \times 20$ dyne
$= 2000 dyne$

AP EAMCET-06.09.2021

Mechanical Properties of Fluids

142723 A body of density $d_{1}$ is counter poised by $M g$ of weights of density $d_{2}$ in air of density $d$ . Then the true mass of the body is

1

M

2

M (\frac{d_{2} - d}{d_{2}})

3

M (\frac{d_{1} - d}{d_{1}})

4

\frac{{Md}_{1} (d_{2} - d)}{d_{2} (d_{1} - d)}

Explanation:

D Apparent Weight (A.W) = True Weight Weight of liquid displaced
$AW = Mg - \frac{M}{d_{2}} \cdot dg$
$AW = M_{o} g - \frac{M_{o}}{d_{1}} \cdot dg$
From equation (i) and (ii), we get -
$M g - \frac{M}{d_{2}} d g = M_{o} g - \frac{M_{o}}{d_{1}} d g$
$\Rightarrow M_{o} (1 - \frac{d}{d_{1}}) = M (1 - \frac{d}{d_{2}})$
$\Rightarrow M_{o} = \frac{M (1 - \frac{d}{d_{2}})}{1 - \frac{d}{d_{1}}}$
$\Rightarrow M_{o} = \frac{M_{1} (d_{2} - d)}{d_{2} (d_{1} - d)}$

AP EAMCET-24.09.2020

Mechanical Properties of Fluids

142725 A metal plate of area $100 {cm}^{2}$ is lying on a liquid layer of thickness $2 mm$ . If the coefficient of viscosity of the liquid is $2 N$ . $s^{- 2} m^{- 2}$ , then find the minimum horizontal force required to move the plate with a speed of $1 cm . s^{- 1}$

1

0.1 N

2

1 N

3

0.5 N

4

0.25 N

Explanation:

A Given that,
Area (A) $= 100 {cm}^{2} = 100 \times 10^{- 4} m^{2}$
Thickness (dy) $= 2 mm = 2 \times 10^{- 3} m$
Coefficient of viscosity $= 2$ N.s.m $^{- 2}$
$Velocity = 1 cm s^{- 1} = 1 \times 10^{- 2} {ms}^{- 1}$
$F = η A (\frac{dv}{dy})$
$∴ F = \frac{2 \times 100 \times 10^{- 4} \times 1 \times 10^{- 2}}{2 \times 10^{- 3}} = 10^{- 1}$
$F = \frac{1}{10} = 0.1 N$

AP EAMCET (Medical)-07.10.2020

Mechanical Properties of Fluids

142727 Find the Young's modulus of the wire whose stress-strain curve is as shown in the following figure:

1

8 \times 10^{11} N . m^{- 2}

2

24 \times 10^{11} N \cdot m^{- 2}

3

10 \times 10^{11} N \cdot m^{- 2}

4

2 \times 10^{11} N \cdot m^{- 2}

Explanation:

D From above graph it is clear that stress $=$ $8 \times 10^{7} N / m^{2}$ and strain $= 4 \times 10^{- 4}$
We know that,
$Young's Modulus (Y) = \frac{Stress}{Strain}$
$Y = \frac{8 \times 10^{7}}{4 \times 10^{- 4}} = 2 \times 10^{11} N / m^{2}$

AP EAMCET-24.08.2021

Mechanical Properties of Fluids

142728 An air bubble of radius $1.0 cm$ rises with a constant speed of $3.5 mm s^{- 1}$ through a liquid of density $1.75 \times 10^{3} {kgm}^{- 3}$ . Neglecting the density of air, the coefficient of viscosity of the liquid is ${k g m}^{- 1} s^{- 1}$

1 54.5

2 109

3 163.5

4 218

Explanation:

B Given that,
Radius of air bubble $= 1.0 cm = 1 \times 10^{- 2} m$
Speed $(v) = 3.5 mm / s = 3.5 \times 10^{- 3} m / s$
Density $(ρ) = 1.75 \times 10^{3} kg / m^{3}$
Coefficient of viscosity $(η) =$ ?
We know that,
Terminal velocity $= \frac{2 r^{2} (ρ) g}{9 η}$
$3.5 \times 10^{- 3} = \frac{2 \times 10^{- 4} \times 1.75 \times 10^{3} \times 9.8}{9 \times η}$
$η = \frac{2 \times 10^{- 4} \times 1.75 \times 10^{3} \times 9.8}{9 \times 3.5 \times 10^{- 3}}$
$η = 108.88 ≃ 109 {kgm}^{- 1} s^{- 1}$

AP EAMCET-03.09.2021

Mechanical Properties of Fluids

142729 A flat plate of area $10 {cm}^{2}$ is separated and a large plate by a layer of Glycerine $1 mm$ thick. If the coefficient of viscosity of Glycerine is $20$ poise. The force required to keep the plate moving the velocity of $1 cm . s^{- 1}$ is

1 80 dyne

2 200 dyne

3 800 dyne

4 2000 dyne

Explanation:

D Given that,
$Area (A) = 10 {cm}^{2}$
Coefficient of viscosity of Glycerine $= 20$ poise Velocity $= 1 cm / s$
We know that,
Force required $= η A \frac{dv}{dy}$
Now $\frac{dv}{dy} = \frac{1 {cms}^{- 1}}{1 \times 10^{- 1} cm} = 10 s^{- 1}$
So, force to keep plate moving $= 10 \times 10 \times 20$ dyne
$= 2000 dyne$

AP EAMCET-06.09.2021

Mechanical Properties of Fluids

142723 A body of density $d_{1}$ is counter poised by $M g$ of weights of density $d_{2}$ in air of density $d$ . Then the true mass of the body is

1

M

2

M (\frac{d_{2} - d}{d_{2}})

3

M (\frac{d_{1} - d}{d_{1}})

4

\frac{{Md}_{1} (d_{2} - d)}{d_{2} (d_{1} - d)}

Explanation:

D Apparent Weight (A.W) = True Weight Weight of liquid displaced
$AW = Mg - \frac{M}{d_{2}} \cdot dg$
$AW = M_{o} g - \frac{M_{o}}{d_{1}} \cdot dg$
From equation (i) and (ii), we get -
$M g - \frac{M}{d_{2}} d g = M_{o} g - \frac{M_{o}}{d_{1}} d g$
$\Rightarrow M_{o} (1 - \frac{d}{d_{1}}) = M (1 - \frac{d}{d_{2}})$
$\Rightarrow M_{o} = \frac{M (1 - \frac{d}{d_{2}})}{1 - \frac{d}{d_{1}}}$
$\Rightarrow M_{o} = \frac{M_{1} (d_{2} - d)}{d_{2} (d_{1} - d)}$

AP EAMCET-24.09.2020

Mechanical Properties of Fluids

142725 A metal plate of area $100 {cm}^{2}$ is lying on a liquid layer of thickness $2 mm$ . If the coefficient of viscosity of the liquid is $2 N$ . $s^{- 2} m^{- 2}$ , then find the minimum horizontal force required to move the plate with a speed of $1 cm . s^{- 1}$

1

0.1 N

2

1 N

3

0.5 N

4

0.25 N

Explanation:

A Given that,
Area (A) $= 100 {cm}^{2} = 100 \times 10^{- 4} m^{2}$
Thickness (dy) $= 2 mm = 2 \times 10^{- 3} m$
Coefficient of viscosity $= 2$ N.s.m $^{- 2}$
$Velocity = 1 cm s^{- 1} = 1 \times 10^{- 2} {ms}^{- 1}$
$F = η A (\frac{dv}{dy})$
$∴ F = \frac{2 \times 100 \times 10^{- 4} \times 1 \times 10^{- 2}}{2 \times 10^{- 3}} = 10^{- 1}$
$F = \frac{1}{10} = 0.1 N$

AP EAMCET (Medical)-07.10.2020

Mechanical Properties of Fluids

142727 Find the Young's modulus of the wire whose stress-strain curve is as shown in the following figure:

1

8 \times 10^{11} N . m^{- 2}

2

24 \times 10^{11} N \cdot m^{- 2}

3

10 \times 10^{11} N \cdot m^{- 2}

4

2 \times 10^{11} N \cdot m^{- 2}

Explanation:

D From above graph it is clear that stress $=$ $8 \times 10^{7} N / m^{2}$ and strain $= 4 \times 10^{- 4}$
We know that,
$Young's Modulus (Y) = \frac{Stress}{Strain}$
$Y = \frac{8 \times 10^{7}}{4 \times 10^{- 4}} = 2 \times 10^{11} N / m^{2}$

AP EAMCET-24.08.2021

Mechanical Properties of Fluids

142728 An air bubble of radius $1.0 cm$ rises with a constant speed of $3.5 mm s^{- 1}$ through a liquid of density $1.75 \times 10^{3} {kgm}^{- 3}$ . Neglecting the density of air, the coefficient of viscosity of the liquid is ${k g m}^{- 1} s^{- 1}$

1 54.5

2 109

3 163.5

4 218

Explanation:

B Given that,
Radius of air bubble $= 1.0 cm = 1 \times 10^{- 2} m$
Speed $(v) = 3.5 mm / s = 3.5 \times 10^{- 3} m / s$
Density $(ρ) = 1.75 \times 10^{3} kg / m^{3}$
Coefficient of viscosity $(η) =$ ?
We know that,
Terminal velocity $= \frac{2 r^{2} (ρ) g}{9 η}$
$3.5 \times 10^{- 3} = \frac{2 \times 10^{- 4} \times 1.75 \times 10^{3} \times 9.8}{9 \times η}$
$η = \frac{2 \times 10^{- 4} \times 1.75 \times 10^{3} \times 9.8}{9 \times 3.5 \times 10^{- 3}}$
$η = 108.88 ≃ 109 {kgm}^{- 1} s^{- 1}$

AP EAMCET-03.09.2021

Mechanical Properties of Fluids

142729 A flat plate of area $10 {cm}^{2}$ is separated and a large plate by a layer of Glycerine $1 mm$ thick. If the coefficient of viscosity of Glycerine is $20$ poise. The force required to keep the plate moving the velocity of $1 cm . s^{- 1}$ is

1 80 dyne

2 200 dyne

3 800 dyne

4 2000 dyne

Explanation:

D Given that,
$Area (A) = 10 {cm}^{2}$
Coefficient of viscosity of Glycerine $= 20$ poise Velocity $= 1 cm / s$
We know that,
Force required $= η A \frac{dv}{dy}$
Now $\frac{dv}{dy} = \frac{1 {cms}^{- 1}}{1 \times 10^{- 1} cm} = 10 s^{- 1}$
So, force to keep plate moving $= 10 \times 10 \times 20$ dyne
$= 2000 dyne$

AP EAMCET-06.09.2021

Mechanical Properties of Fluids

142723 A body of density $d_{1}$ is counter poised by $M g$ of weights of density $d_{2}$ in air of density $d$ . Then the true mass of the body is

1

M

2

M (\frac{d_{2} - d}{d_{2}})

3

M (\frac{d_{1} - d}{d_{1}})

4

\frac{{Md}_{1} (d_{2} - d)}{d_{2} (d_{1} - d)}

Explanation:

D Apparent Weight (A.W) = True Weight Weight of liquid displaced
$AW = Mg - \frac{M}{d_{2}} \cdot dg$
$AW = M_{o} g - \frac{M_{o}}{d_{1}} \cdot dg$
From equation (i) and (ii), we get -
$M g - \frac{M}{d_{2}} d g = M_{o} g - \frac{M_{o}}{d_{1}} d g$
$\Rightarrow M_{o} (1 - \frac{d}{d_{1}}) = M (1 - \frac{d}{d_{2}})$
$\Rightarrow M_{o} = \frac{M (1 - \frac{d}{d_{2}})}{1 - \frac{d}{d_{1}}}$
$\Rightarrow M_{o} = \frac{M_{1} (d_{2} - d)}{d_{2} (d_{1} - d)}$

AP EAMCET-24.09.2020

Mechanical Properties of Fluids

142725 A metal plate of area $100 {cm}^{2}$ is lying on a liquid layer of thickness $2 mm$ . If the coefficient of viscosity of the liquid is $2 N$ . $s^{- 2} m^{- 2}$ , then find the minimum horizontal force required to move the plate with a speed of $1 cm . s^{- 1}$

1

0.1 N

2

1 N

3

0.5 N

4

0.25 N

Explanation:

A Given that,
Area (A) $= 100 {cm}^{2} = 100 \times 10^{- 4} m^{2}$
Thickness (dy) $= 2 mm = 2 \times 10^{- 3} m$
Coefficient of viscosity $= 2$ N.s.m $^{- 2}$
$Velocity = 1 cm s^{- 1} = 1 \times 10^{- 2} {ms}^{- 1}$
$F = η A (\frac{dv}{dy})$
$∴ F = \frac{2 \times 100 \times 10^{- 4} \times 1 \times 10^{- 2}}{2 \times 10^{- 3}} = 10^{- 1}$
$F = \frac{1}{10} = 0.1 N$

AP EAMCET (Medical)-07.10.2020

Mechanical Properties of Fluids

142727 Find the Young's modulus of the wire whose stress-strain curve is as shown in the following figure:

1

8 \times 10^{11} N . m^{- 2}

2

24 \times 10^{11} N \cdot m^{- 2}

3

10 \times 10^{11} N \cdot m^{- 2}

4

2 \times 10^{11} N \cdot m^{- 2}

Explanation:

D From above graph it is clear that stress $=$ $8 \times 10^{7} N / m^{2}$ and strain $= 4 \times 10^{- 4}$
We know that,
$Young's Modulus (Y) = \frac{Stress}{Strain}$
$Y = \frac{8 \times 10^{7}}{4 \times 10^{- 4}} = 2 \times 10^{11} N / m^{2}$

AP EAMCET-24.08.2021

Mechanical Properties of Fluids

142728 An air bubble of radius $1.0 cm$ rises with a constant speed of $3.5 mm s^{- 1}$ through a liquid of density $1.75 \times 10^{3} {kgm}^{- 3}$ . Neglecting the density of air, the coefficient of viscosity of the liquid is ${k g m}^{- 1} s^{- 1}$

1 54.5

2 109

3 163.5

4 218

Explanation:

B Given that,
Radius of air bubble $= 1.0 cm = 1 \times 10^{- 2} m$
Speed $(v) = 3.5 mm / s = 3.5 \times 10^{- 3} m / s$
Density $(ρ) = 1.75 \times 10^{3} kg / m^{3}$
Coefficient of viscosity $(η) =$ ?
We know that,
Terminal velocity $= \frac{2 r^{2} (ρ) g}{9 η}$
$3.5 \times 10^{- 3} = \frac{2 \times 10^{- 4} \times 1.75 \times 10^{3} \times 9.8}{9 \times η}$
$η = \frac{2 \times 10^{- 4} \times 1.75 \times 10^{3} \times 9.8}{9 \times 3.5 \times 10^{- 3}}$
$η = 108.88 ≃ 109 {kgm}^{- 1} s^{- 1}$

AP EAMCET-03.09.2021

Mechanical Properties of Fluids

142729 A flat plate of area $10 {cm}^{2}$ is separated and a large plate by a layer of Glycerine $1 mm$ thick. If the coefficient of viscosity of Glycerine is $20$ poise. The force required to keep the plate moving the velocity of $1 cm . s^{- 1}$ is

1 80 dyne

2 200 dyne

3 800 dyne

4 2000 dyne

Explanation:

D Given that,
$Area (A) = 10 {cm}^{2}$
Coefficient of viscosity of Glycerine $= 20$ poise Velocity $= 1 cm / s$
We know that,
Force required $= η A \frac{dv}{dy}$
Now $\frac{dv}{dy} = \frac{1 {cms}^{- 1}}{1 \times 10^{- 1} cm} = 10 s^{- 1}$
So, force to keep plate moving $= 10 \times 10 \times 20$ dyne
$= 2000 dyne$

AP EAMCET-06.09.2021

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