Tag: viscosity

Questions Related to viscosity

A water hose 2 cm in diameter is used to fill a 20 litre bucket. If it takes 1 minute to fill bucket with watch velocity it leaves the hose ,

viscosity option b: engineering physics properties of matter physics

  1. 150 cm/s

  2. 70 cm/s

  3. 106 cm/s

  4. 100 cm/s

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Correct Option: C
Explanation:

Given,

Area, $A=\dfrac{\pi {{d}^{2}}}{4}=\dfrac{\pi }{4}{{\left( 0.02 \right)}^{2}}$

Volume rate of flow, $\dot{V}=\dfrac{volume}{time}=\dfrac{20\,L}{60}=\dfrac{{{10}^{-3}}}{3}\,{{m}^{3}}{{s}^{-1}}$

Volume rate of flow = Cross-Section Area x Velocity of Flow

$ \dot{V}=Av $

$v=\dfrac{{\dot{V}}}{A}=\dfrac{{{10}^{-3}}}{3}\times \dfrac{4}{\pi {{\left( 0.02 \right)}^{2}}}=1.06\,m{{s}^{-1}}=106\,cm{{s}^{-1}}$

Hence, velocity of water leaves hose is$106\,cm{{s}^{-1}}$.

An air bubble of diameter 2mm rises steadily througha solution of density $1750 kg/m^3$at the rate of $0.35cm/s$.Calculate the coefficient of viscosity of the solution.The density of air is negligible. 

viscosity option b: engineering physics properties of matter physics

  1. 10

  2. 11

  3. 12

  4. 13

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Correct Option: B
Explanation:

The force of buoyancy B is equal to the weight of  the displaced liquid. Thus

$\Rightarrow B=\dfrac{4}{3}\pi r^36g$
This force is upward. The viscous force acting downward is $F=6\pi nrv$
The weight of the air bubble may be neglected as the density of air is small. for uniform velocity -
$\Rightarrow F=B$
$\Rightarrow 6\pi nrv=\dfrac{4}{3}\pi r^3 6g$
$\Rightarrow n=\dfrac{2r^36g}{9v}$
         $=\dfrac{2\times \left( 1\times 106{-3}m\right)^2\times \left( 1750kg/m^3\right)\times 9.8m/s^2}{9\times 0.35\times 10^{-2}m/s}$
         $=11\;poise$
This appears to be a highly viscous liquid.

Blood vessel is $0.10\ m$ in length and has a radius of $1.5\times{10}^{-3}m$. Blood flows at rate of ${10}^{-7}{m}^{-3}/s$ through this vessel. The pressure difference that must be maintained in this flow, between the two ends of the vessel is $20\ Pa$. What is the viscosity sufficient of blood?

viscosity option b: engineering physics properties of matter physics

  1. $2\times{10}^{-3}\ Pa-s$

  2. $1\times{10}^{-3}\ Pa-s$

  3. $4\times{10}^{-3}\ Pa-s$

  4. $5\times{10}^{-4}\ Pa-s$

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Correct Option: C

A U-tube having identical limbs is partially filled with water. An immiscible oil having a density of 0.8 g/cc is poured into one side until the water rises by 25 cm on the other side. the level of oil will stand higher than the water level? 

viscosity option b: engineering physics properties of matter physics

  1. 6.25 cm

  2. 75 cm

  3. 22.5 cm

  4. 12.5 cm

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Correct Option: A

A small sphere of mass M and density $D _1$ is dropped in a vessel filled with glycerine. If the density of glycerine is $D _2$ then the viscous force acting on the ball will be in Newton.

viscosity option b: engineering physics properties of matter physics

  1. $M D _1 D _2$

  2. $Mg \displaystyle \left [ 1- \frac {D _2}{D _1} \right ]$

  3. $\displaystyle \frac {M D _1 g}{D _2}$

  4. $\displaystyle \frac {M}{g} \left ( D _1 + D _2 \right)$

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Correct Option: B
Explanation:

When sphere is in glycerine, three forces acts on it. which balances each other.

  • weight $(W)$
  • buoyant force $(F _B)$
  • viscous force $(F _V)$
and $W=F _B+F _V$
$\Rightarrow Mg=V _g D _2g +F _V$
$\Rightarrow Mg=V _s D _2g +F _V           \because V _g=V _s$
$\Rightarrow Mg=\frac{M}{D _1} D _2g +F _V             \because V _s=\frac{M}{D _1}$
$\Rightarrow F _V=Mg \left [ 1- \frac {D _2}{D _1} \right ]$

The viscous drag on a spherical body moving with a speed V is proportional to:

viscosity option b: engineering physics properties of matter physics

  1. $\sqrt V$

  2. $V$

  3. $\displaystyle \frac{1}{\sqrt V}$

  4. $V^{2}$

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Correct Option: B
Explanation:

The viscous drag on a spherical body is given as $F=6\pi \eta RV$. Here $\eta$ is the coefficient of viscosity, R is the radius of the sphere and V is its velocity. 

An air bubble of radius $1 \,cm$ is found to rise in a cylindrical vessel of large radius at a steady rate of $0.2 \,cm$ per second. If the density of the liquid is $1470 \,kg \,m^{-3}$, then coefficient of viscosity of liquid is approximately equal to

viscosity option b: engineering physics properties of matter physics

  1. $163$ poise

  2. $163$ centi-poise

  3. $140$ poise

  4. $140$ centi-poise

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Correct Option: B

A capillary tube of area of cross-section A is dipped in water vertically. The amount of heat evolved as the water rises in the capillary tube up to height h is: (The density of water is $\rho$)

viscosity option b: engineering physics properties of matter physics

  1. $\dfrac{A\rho gh^2}{2}$

  2. $Agh^2\rho$

  3. $2Agh^2\rho$

  4. None of these

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Correct Option: A

Viscous force is somewhat like friction as it opposes, the motion and is non-conservative but not exactly so, because

viscosity option b: engineering physics properties of matter physics

  1. It is velocity dependent while friction is not

  2. It is velocity independent while friction is not

  3. It is temperature dependent while friction is not

  4. It is independent of area is like surface tension while friction is dependent

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Correct Option: A,C
Explanation:

Using the relation $\tau=\mu\displaystyle\frac{du}{dy}$ we see that viscosity is velocity dependent. Also viscosity decreases with increasing temperature. Thus A and C are correct.

A liquid flows between two parallel plates along the x-axis. The difference between the velocity of two  layers separated by the distance $dy$ is $dv$. If $A$ is the area of each plate, then Newton's law of viscosity may be written as:

viscosity option b: engineering physics properties of matter physics

  1. $F=-\eta A\dfrac{dv}{dx}$

  2. $F=+\eta A\dfrac{dv}{dx}$

  3. $F=-\eta A\dfrac{dv}{dy}$

  4. $F=+\eta A\dfrac{dv}{dy}$

Show answer
Correct Option: C
Explanation:

The Newton's viscous force acting between two liquid surfaces with relative velocity $dv$ and distance $dy$ between the layers is given as $-\eta A\dfrac{dv}{dy}$