Question Set

1. According to Newton's law of viscosity, the shear stress on a layer of a fluid is __________ to the rate of shear strain.




Ask Your Doubts Here

Type in
(Press Ctrl+g to toggle between English and the chosen language)

Comments

Show Similar Question And Answers
QA->The process of transfer of heat from one point of a fluid to another part of a fluid by the movement of fluid itself is called?....
QA->Decrease of viscosity with increasing shear rate is known as....
QA->According to Newton’s third law every force is accompanied by an equal and opposite force. Then how can a movement take place?....
QA->Name the economy in which people work according to the principle ‘from each according to his ability to each according to his needs’?....
QA->Viscosity of gas is less than viscosity of?....
MCQ->According to Newton's law of viscosity, the shear stress on a layer of a fluid is __________ to the rate of shear strain.....
MCQ->As per Newton's law of viscosity, the shear stress for a given rate of angular deformation of fluid is proportional to(where, μ = fluid viscosity)....
MCQ->Which of the following must be followed by the flow of a fluid (real or ideal)? (I) Newton's law of viscosity. (II) Newton's second law of motion. (III) The continuity equation. (IV) Velocity of boundary layer must be zero relative to boundary. (V) Fluid cannot penetrate a boundary.....
MCQ->In Newton's law of viscosity, which states that the shear stress is proportional to the __________ Co-efficient of viscosity is called dynamic or absolute viscosity(where, V = velocity, Vg = velocity gradient ).....
MCQ->In a plane strain problem in XY plane, the shear strain = 12 x 10-6, and the normal strain in X and Y direction = 0, For this state of strain, what is the diameter of the Mohr's Circle of strain ?....
Terms And Service:We do not guarantee the accuracy of available data ..We Provide Information On Public Data.. Please consult an expert before using this data for commercial or personal use | Powered By:Omega Web Solutions
© 2002-2017 Omega Education PVT LTD...Privacy | Terms And Conditions
Question ANSWER With Solution