Viscosity MCQ Quiz in मल्याळम - Objective Question with Answer for Viscosity - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Mar 12, 2025
Latest Viscosity MCQ Objective Questions
Top Viscosity MCQ Objective Questions
Viscosity Question 1:
Viscosity has dimension of
Answer (Detailed Solution Below)
Viscosity Question 1 Detailed Solution
Concept:
Viscosity: The property that represents the internal resistance of a fluid to motion (i.e. fluidity) is called viscosity. There are two ways to write viscosities:
- Dynamic viscosity: This is also termed as absolute viscosity. A common unit of dynamic viscosity is poise.
- 1 poise = 0.1 Pa.s = 0.1 Ns/m2 = kg/ms
- Kinematic Viscosity: The ratio of dynamic viscosity to density appears frequently and this ratio is given by the name kinematic viscosity. Its unit is Stoke or m2/s (1 stoke = 0.0001 m2 /s).
- Viscosity is usually termed to denote dynamic viscosity.
- Unit of μ = \(\frac{kg}{ms}=\frac{M}{LT}=M^{1}L^{-1}T^{-1}\)
Viscosity Question 2:
An oil of specific gravity 0.9 has viscosity of 0.28 Stokes at 38°C. What will be its viscosity in Ns/m2 ?
Answer (Detailed Solution Below)
Viscosity Question 2 Detailed Solution
Explanation:
- Viscosity can be defined as the measure of a fluid's resistance to deformation at a given flow.
- Caused by friction within a fluid.
- Result of intermolecular forces between the particles within a fluid.
- SI unit N-s/m2 or Pa-s.
- There are two kinds of viscosity:
- Dynamic Viscosity ( µ )
- SI unit = Ns / m2 or Pa-s
- CGS unit = Poise
- MKS unit = \(\dfrac{kgf - \sec}{m^2}\)
- 1 Poise = 0.1 Pa-s
- Kinematic Viscosity \((\upsilon\;)\)
- \(ν =\frac{μ}{ρ}\)
- MKS unit = SI unit = m2/s
- CGS unit = Stokes or cm2/s
- 1 Stoke = 10-4 m2/s
Calculation:
Given:
ρ = 900 kg/m3, ν = 0.28 × 10-4 m2/s
\(ν =\frac{μ}{ρ}\)
μ = 900 × 0.28 × 10-4
μ = 0.0252 Pa-s
Viscosity Question 3:
Kinematic viscosity of air at 20°C is given to be 1.6 × 10 -5 m2 /s. Its kinematic viscosity at 70°C will be approximately.
Answer (Detailed Solution Below)
Viscosity Question 3 Detailed Solution
Concept:
Variation of viscosity with temperature:
Dynamic viscosity is directly proportional to the square-root of Absolute temperature.
i.e. \(\mu \propto \sqrt T \;\)
Variation of density with temperature:
PV = mRT ⇒ P = ρRT
\(\rho=\frac{P}{RT} ⇒ \rho\propto\frac{1}{T}\)
Kinematic viscosity (ν):
It is the ratio of dynamic viscosity to the density of the fluid.
\(ν=\frac{\mu}{\rho}\)
\(\thereforeν\propto\sqrt{T}\;and \;ν\propto\frac{1}{\frac{1}{T}}\)
\(\therefore ν\propto T^{3/2}\)
Calculation:
Given:
T1 = 20°C ⇒ 273 + 20 = 293 K, ν1 = 1.6 × 10-5 m2/s, T2 = 70°C ⇒ 343 K.
\(ν\propto T^{3/2}\)
\(\frac{ν_1}{T_1^{3/2}} =\frac{ν_2}{T_2^{3/2}} \)
\(\nu_2=\nu_1\;\times\;\left(\frac{T_2}{T_1}\right)^{3/2}\)
\(\nu_2=1.6\;\times\;10^{-5}\;\times\;\left(\frac{343}{293}\right)^{3/2}\)
\(\nu_2=2.02\;\times\;10^{-5}\;\;m^2/s\)
Viscosity Question 4:
For a Newtonian Fluid:
Answer (Detailed Solution Below)
Viscosity Question 4 Detailed Solution
CONCEPT:
Newton's law of Viscous Force:
- According to Newton's hypothesis, the tangential force ‘F’ acting on a plane parallel layer is proportional to the area of the plane ‘A’ and the velocity gradient \(\frac{{dv}}{{dx}}\) in a direction normal to the layer,
\(F = - \eta A\frac{{dv}}{{dx}}\)
Where η = coefficient of viscosity, A = area of the plane, and dv/dx = velocity gradient.
- A negative sign is employed because viscous force acts in a direction opposite to the flow of liquid.
- The SI unit of viscosity is poiseiulle (Pl). Its other units are Nsm-2 or Pa s.
EXPLANATION:
- Newtonian fluids defined as fluids for which the shear stress is linearly proportional to the shear strain rate. Therefore the shear stress-strain graph for a Newtonian fluid is a straight line. Hence, option 1 is correct. The fluids, such as air, kerosene, gasoline, and other oil-based liquids, are Newtonian fluids.
- Newtonian fluids are analogous to elastic solids (Hooke’s law: stress proportional to strain).
- Fluids for which the shear stress is not linearly related to the shear strain rate are called non- Newtonian fluids. Examples include slurries, colloidal suspensions, polymer solutions, blood, paste, and cake batter.
Viscosity Question 5:
MKS unit of viscosity is -
Answer (Detailed Solution Below)
Viscosity Question 5 Detailed Solution
Explanation:
- Viscosity can be defined as the measure of a fluid's resistance to deformation at a given flow.
- Caused by friction within a fluid.
- Result of intermolecular forces between the particles within a fluid.
- SI unit N-s/m2 or Pa-s.
- Kinematic viscosity = Dynamic viscosity / Density of fluid
Shear stress is given by,
\(Shear\;stress = \;\mu \;\frac{{du}}{{dy}}\)
where,
µ = Viscosity of fluid
\(\frac{u}{y}\)= Rate of shear deformation
There are two kinds of viscosity:
1. Dynamic Viscosity ( µ )
- SI unit = Ns / m2 or Pa-s
- CGS unit = Poise
- MKS unit = \(\dfrac{kgf - \sec}{m^2}\)
- 1 Poise = 0.1 Pa-s
2. Kinematic Viscosity \((\upsilon\;)\)
- MKS unit = SI unit = m2/s
- CGS unit = Stokes or cm2/s
- 1 Stoke = 10-4 m2/s
Viscosity Question 6:
The dynamic viscosity (poise) of a liquid having kinematic viscosity 6 stokes and specific gravity 2.0 is __________.
Answer (Detailed Solution Below) 12
Viscosity Question 6 Detailed Solution
Concept:
Specific gravity = ρliquid/ρwater
Kinematic viscosity (ν) = μ/ρ
Dynamic viscosity (μ) = νρ
where, ρ = density
Calculation:
Given:
ν = 6 cm2/s = 6 × 10-4 m2/s
S.G. = 2 = ρ/ρw
⇒ ρ = SG × 1000 = 2000 kg/m3
Now,
μ = 6 × 10-4 × 2000 = 1.2 Ns/m2 = 12 poise
Important conversions:
1 Poise = 0.1 Pa.s
1 stoke = 1 cm2/s = 0.0001 m2/s
Viscosity Question 7:
Increase of temperature
Answer (Detailed Solution Below)
Viscosity Question 7 Detailed Solution
Concept:
Viscosity: The property that represents the internal resistance of a fluid to motion (i.e. fluidity) is called viscosity.
There are two ways to write viscosity:
- Dynamic viscosity: This is also termed absolute viscosity. A common unit of dynamic viscosity is poise
1 poise = 0.1 Pa.s = 0.1 N.s/m2
- Kinematic Viscosity: The ratio of dynamic viscosity to density appears frequently and this ratio is given by the name kinematic viscosity. Its unit is Stoke or m2/s (1 stoke = 0.0001 m2 /s).
Viscosity is affected by temperature.
- The viscosity of liquids decreases but that of gases increases with an increase in temperature.
- This is due to the reason that in liquids, the shear stress is due to the inter-molecular cohesion which decreases with the increase of temperature.
- In gases, the inter-molecular cohesion is negligible and the shear stress is due to the exchange of momentum of the molecules, normal to the direction of motion.
- The molecular activities increase with rising temperature and so the viscosity of gases increases.
Viscosity Question 8:
In which types of fluids, it is observed that momentum transfer dominates cohesive forces with an increase in temperature, and hence viscosity increases?
Answer (Detailed Solution Below)
Viscosity Question 8 Detailed Solution
Concept:
Viscosity:
- The property of fluid (a liquid or a gas) to resist any flow is called viscosity.
- The upper layer of fluid causes shear stress on the adjacent lower layer and the lower layer in return causes shear stress on the adjacent upper layer.
- Viscous force in a fluid is due to cohesive force and molecular momentum transfer.
- Cohesive Force: The force of attraction between the molecules of the same substance.
- Molecular Moment transfer: The phenomena in which the molecules of a fluid drag the surrounding molecules to move along with them.
Viscosity relation with temperature:
Liquid:
- In liquids, the main cause of viscosity is cohesion between the molecules.
- With the increase in temperature, this cohesive force decreases as the distance between the molecules become more due to the increase in energy of particles hence the movement of particles becomes easy.
- Hence viscosity of liquid decreases with an increase in temperature.
Gases:
- In gases, the important cause of viscosity is randomness i.e. molecular momentum transfer.
- Due to the rise in temperature, the kinetic energy of molecules increases which makes Crms increase, hence randomness and collision of molecules increased.
- Hence viscosity of gases increases with an increase in temperature.
Viscosity Question 9:
\(\nu = 0.0022t - \frac{{1.8}}{t}\) is the equation to determine kinematic viscosity of liquids by
Answer (Detailed Solution Below)
Viscosity Question 9 Detailed Solution
The device used for the measurement of viscosity is known as viscometer. According to Saybolt viscometer:
\(\nu = \frac{\mu }{\rho } = 0.0022t - \frac{{1.8}}{t}\)Viscosity Question 10:
Which of the following statements are true?
Answer (Detailed Solution Below)
Viscosity Question 10 Detailed Solution
Explanation:
Viscosity: The property that represents the internal resistance of a fluid to motion (i.e. fluidity) is called viscosity
The viscosity of liquids decreases with temperature, whereas the viscosity of gases increases with temperature.
For Liquid:
In case of the liquid, the molecules possess more energy at higher temperatures, and they can oppose the large cohesive intermolecular forces more strongly. As a result, the energized liquid molecules can move more freely.
For gases:
In a gas, on the other hand, the intermolecular forces are negligible, and the gas molecules at high temperatures move randomly at higher velocities. This results in more molecular collisions per unit volume per unit time that is more momentum transfer and therefore in greater resistance to flow.
So, As the temperature increases, the coefficient of viscosity of gases increase.