The viscous force acting between two lay...

The viscous force acting between two layers of a liquid is given by `F/A=eta(dv)/(dz)`. This F/A may be called

pressure

longitudinal stress

tangential stress

volume stress

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When a liquid flows in a tube, there is relative motion between adjacent layers of the liquid. This force is called the viscous force which tends to oppose the relative motion between the layers of the liquid. Newton was the first person to study the factors that govern the viscous force in a liquid. According to Newton’s law of viscous flow, the magnitude of the viscous force on a certain layer of a liquid is given by F = - eta A (dv)/(dx) where A is the area of the layer (dv)/(dx) is the velocity gradient at the layer and eta is the coefficient of viscosity of the liquid. If f is the frictional force between a solid sliding over another solid, and F is the viscous force when a liquid layer slides over another, then :

When a liquid flows in a tube, there is relative motion between adjacent layers of the liquid. This force is called the viscous force which tends to oppose the relative motion between the layers of the liquid. Newton was the first person to study the factors that govern the viscous force in a liquid. According to Newton’s law of viscous flow, the magnitude of the viscous force on a certain layer of a liquid is given by F = - eta A (dv)/(dx) where A is the area of the layer (dv)/(dx) is the velocity gradient at the layer and eta is the coefficient of viscosity of the liquid. The dimensional formula for the coefficient of viscosity is :

When a liquid flows in a tube, there is relative motion between adjacent layers of the liquid. This force is called the viscous force which tends to oppose the relative motion between the layers of the liquid. Newton was the first person to study the factors that govern the viscous force in a liquid. According to Newton’s law of viscous flow, the magnitude of the viscous force on a certain layer of a liquid is given by F = - eta A (dv)/(dx) where A is the area of the layer (dv)/(dx) is the velocity gradient at the layer and eta is the coefficient of viscosity of the liquid. A river is 5 m deep. The velocity of water on its surface is 2 ms^(-1) If the coefficient of viscosity of water is 10 ^(-3 ) Nsm ^(-2) , the viscous force per unit area is :

According to Newton, the viscous force acting between liquid layers of area A and velocity gradient (Deltav)/(Deltaz) is given by F =- eta A (dv)/(dz) , where eta is constant called

Frictional force between solids operates even when they do not move with respect to each other. Do we have viscous force acting between two layers even if there is no relative motion?

When a viscous liquid flows , adjacent layers oppose their relative motion by applying a viscous force given by F = - eta A (dv)/(dz) where , ete = coefficient of viscosity , A = surface area of adjacent layers in contact , (dv)/(dz) = velocity gradient Now , a viscous liquid having coefficient of viscosity eta is flowing through a fixed tube of length l and radius R under a pressure difference P between the two ends of the tube . Now consider a cylindrical vloume of liquid of radius r . Due to steady flow , net force on the liquid in cylindrical volume should be zero . - eta 2pirl (dv)/(dr) = Ppir^(2) - int _(v)^(0),dv = P/(2 eta l) int_(tau)^(R) rdr ( :' layer in contact with the tube is stationary ) v = v_(0) (1- (r^(2))/(R^(2))) , where v_(0) = (PR^(2))/(4nl) :. " " Q = (piPR^(4))/(8etaL) This is called Poisecuille's equation . The viscous force on the cylindrical volume of the liquid varies as

When a viscous liquid flows , adjacent layers oppose their relative motion by applying a viscous force given by F = - eta A (dv)/(dz) where , eta = coefficient of viscosity , A = surface area of adjacent layers in contact , (dv)/(dz) = velocity gradient Now , a viscous liquid having coefficient of viscosity eta is flowing through a fixed tube of length l and radius R under a pressure difference P between the two ends of the tube . Now consider a cylindrical vloume of liquid of radius r . Due to steady flow , net force on the liquid in cylindrical volume should be zero . - eta 2pirl (dv)/(dr) = Ppir^(2) - int _(v)^(0),dv = P/(2 eta l) int_(tau)^(R) rdr ( :' layer in contact with the tube is stationary ) v = v_(0) (1- (r^(2))/(R^(2))) , where v_(0) = (PR^(2))/(4nl) :. " " Q = (piPR^(4))/(8sta l) This is called Poisecuille's equation . The velocity of flow of liquid at r = R/2 is

When a viscous liquid flows , adjacent layers oppose their relative motion by applying a viscous force given by F = - eta A (dv)/(dz) where , ete = coefficient of viscosity , A = surface area of adjacent layers in contact , (dv)/(dz) = velocity gradient Now , a viscous liquid having coefficient of viscosity eta is flowing through a fixed tube of length l and radius R under a pressure difference P between the two ends of the tube . Now consider a cylindrical vloume of liquid of radius r . Due to steady flow , net force on the liquid in cylindrical vloume should be zero . - eta 2pirl (dv)/(dr) = Ppir^(2) - int _(v)^(0),dv = P/(2 eta l) int_(tau)^(R) rdr ( :' layer in contact with the tube is stationary ) v = v_(0) (1- (r^(2))/(R^(2))) , where v_(0) = (PR^(2))/(4nl) :. " " Q = (piPR^(4))/(8sta l) This is called Poisecuille's equation . The volume of the liquid flowing per sec across the cross - section of the tube is .

A liquid whose coefficient of viscosity is eta flows on a horizontal surface. Let dx represent the vertical distance between two layers of liquid and dv represent the difference in the velocities of the two layers. Then the quantity eta (dv//dx ) has the same dimensions as:

Find the viscous drag between the two liquid layers each of area 100 cm^(2) , and having relative velocity 8 cm s^(-1) . The viscosity of the liquid is 0.004 PI and the layers are separated by a distance 4 cm.

HC VERMA ENGLISH-SOME MECHANICAL PROPERTIES OF MATTER-Objective-1

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