Tag: option b: engineering physics

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Questions Related to option b: engineering physics

Multiple choice viscosity option b: engineering physics properties of matter 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?

  1. 1;2

  2. 1;4

  3. 1;8

  4. 4;1
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Terminal velocity v_t is proportional to r^2. The drag force F = 6 * pi * eta * r * v_t. Substituting v_t proportional to r^2, we get F proportional to r * r^2 = r^3. The ratio of forces for radii r and 2r is (r/2r)^3 = 1/8.

Multiple choice viscosity option b: engineering physics properties of matter physics

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

  1. Glycerine

  2. Water

  3. Kerosene

  4. All of them at the same time
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The rate at which a stirred liquid comes to rest depends on its viscosity. Glycerine has the highest viscosity among the given liquids, causing it to dissipate kinetic energy through internal friction most rapidly.

Multiple choice viscosity option b: engineering physics properties of matter physics

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 -----

  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. $None$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Compressibility K = -(1/V) * (dV/dP). Here, dV/V = 0.2 / 200 = 0.001. dP = 2 * 10^8 Pa. K = 0.001 / (2 * 10^8) = 0.5 * 10^-11 = 5 * 10^-12 m^2/N.

Multiple choice viscosity option b: engineering physics properties of matter physics

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 ,

  1. 150 cm/s

  2. 70 cm/s

  3. 106 cm/s

  4. 100 cm/s
Reveal answer Fill a bubble to check yourself
C Correct answer
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$106\,cm{{s}^{-1}}$.

Multiple choice viscosity option b: engineering physics properties of matter physics

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. 

  1. 10

  2. 11

  3. 12

  4. 13
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

Multiple choice viscosity option b: engineering physics properties of matter physics

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?

  1. $2\times{10}^{-3}\ Pa-s$
  2. $1\times{10}^{-3}\ Pa-s$
  3. $4\times{10}^{-3}\ Pa-s$
  4. $5\times{10}^{-4}\ Pa-s$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Poiseuille's law states Q = (pi * P * r^4) / (8 * eta * L). Solving for eta: eta = (pi * P * r^4) / (8 * Q * L). Plugging in P=20, r=1.5*10^-3, Q=10^-7, L=0.1, we get eta = (3.14 * 20 * (1.5*10^-3)^4) / (8 * 10^-7 * 0.1) = 4 * 10^-3 Pa-s.

Multiple choice viscosity option b: engineering physics properties of matter physics

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? 

  1. 6.25 cm

  2. 75 cm

  3. 22.5 cm

  4. 12.5 cm
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

By hydrostatic equilibrium, the pressure exerted by the oil column must balance the pressure exerted by the displaced water column. Using rho_oil * h_oil = rho_water * h_water with rho_oil = 0.8 g/cc and water rising by 25 cm (meaning the water level difference is 25 cm), we find the oil height is 31.25 cm, making it stand 12.5 cm higher than the final water level.

Multiple choice viscosity option b: engineering physics properties of matter physics

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.

  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. $\displaystyle \frac {M}{g} \left ( D _1 + D _2 \right)$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $W=F _B+F _V$
$\Rightarrow Mg=V _g D _2g +F _V$
$\Rightarrow Mg=V _s D _2g +F _V           \because V _g=V _s$
$\Rightarrow Mg=\frac{M}{D _1} D _2g +F _V             \because V _s=\frac{M}{D _1}$
$\Rightarrow F _V=Mg \left [ 1- \frac {D _2}{D _1} \right ]$

Multiple choice viscosity option b: engineering physics properties of matter physics

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

  1. $\sqrt V$
  2. $V$
  3. $\displaystyle \frac{1}{\sqrt V}$
  4. $V^{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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. 

Multiple choice viscosity option b: engineering physics properties of matter physics

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

  1. $163$ poise
  2. $163$ centi-poise
  3. $140$ poise
  4. $140$ centi-poise
Reveal answer Fill a bubble to check yourself
B Correct answer