Tag: option b: engineering physics

Questions Related to option b: engineering physics

Two solid metal balls of radii $r$ $2r$ are falling with their terminal speeds in a viscous liquid.What is the ratio of drag force acting on these two balls?

viscosity option b: engineering physics properties of matter physics

  1. 1;2

  2. 1;4

  3. 1;8

  4. 4;1

Show answer
Correct Option: C

We have three beakers A, B and C containing glycerine, water and kerosene respectively. They are stirred vigorously and placed on the table. The liquid which comes to rest at the earliest is

viscosity option b: engineering physics properties of matter physics

  1. Glycerine

  2. Water

  3. Kerosene

  4. All of them at the same time

Show answer
Correct Option: A

When $200 ml$ of water is subjected to a pressure of $2 \times {10^8}pa,$ the decrease in its volume is $0.2 ml.$ the compressibility of water is -----

viscosity option b: engineering physics properties of matter physics

  1. $5 \times {10^{ - 8}}{m^2}{N^{ - 1}}$

  2. $5 \times {10^{ - 10}}{m^2}{N^{ - 1}}$

  3. $5 \times {10^{ - 12}}{m^2}{N^{ - 1}}$

  4. $None$

Show answer
Correct Option: C

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

Show answer
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

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

Show answer
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

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

Show answer
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. $V^{2}$

Show answer
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

Show answer
Correct Option: B