Home
Class 11
CHEMISTRY
The coefficient of viscosity (eta) of a ...

The coefficient of viscosity `(eta)` of a fluid moving steadily between two surface is given by the formula `(f) = etaA dV//dx` where `f` is the frictional force on the fluid, `A` is the area in the fluid, and `dV//dx` is velocity gradient inside the fluid at that area. The SI unit of viscosity is given as :

A

`kg m^(-1) s^(-1)`

B

`Nm^(-2) s`

C

Nil

D

Newtons

Text Solution

AI Generated Solution

The correct Answer is:
To find the SI unit of viscosity (η), we start from the formula given: \[ F = \eta A \frac{dV}{dx} \] Where: - \( F \) is the frictional force, - \( \eta \) is the coefficient of viscosity, - \( A \) is the area, - \( \frac{dV}{dx} \) is the velocity gradient. ### Step 1: Rearranging the formula to solve for η We can rearrange the formula to isolate η: \[ \eta = \frac{F}{A \frac{dV}{dx}} \] ### Step 2: Identifying the units of each variable Next, we need to identify the SI units of each variable: - The unit of force \( F \) is Newton (N). - The unit of area \( A \) is square meters (m²). - The velocity gradient \( \frac{dV}{dx} \) has units of velocity per unit distance. The unit of velocity \( V \) is meters per second (m/s), and the unit of distance \( x \) is meters (m). Therefore, the unit of \( \frac{dV}{dx} \) is: \[ \frac{m/s}{m} = s^{-1} \] ### Step 3: Substituting the units into the equation for η Now we can substitute the units into the equation for η: \[ \text{Unit of } \eta = \frac{\text{Unit of } F}{\text{Unit of } A \times \text{Unit of } \frac{dV}{dx}} \] Substituting the units we identified: \[ \text{Unit of } \eta = \frac{N}{m^2 \cdot s^{-1}} \] ### Step 4: Converting Newton into base SI units Recall that 1 Newton (N) can be expressed in terms of base SI units: \[ 1 \, N = 1 \, \text{kg} \cdot \text{m/s}^2 \] Thus, we can rewrite the unit of η: \[ \text{Unit of } \eta = \frac{kg \cdot m/s^2}{m^2 \cdot s^{-1}} \] ### Step 5: Simplifying the units Now, we simplify the expression: \[ \text{Unit of } \eta = \frac{kg \cdot m}{s^2} \cdot \frac{s}{m^2} \] This simplifies to: \[ \text{Unit of } \eta = \frac{kg}{m \cdot s} \] ### Conclusion Thus, the SI unit of viscosity (η) is: \[ \text{Unit of } \eta = \text{kg} \cdot \text{m}^{-1} \cdot \text{s}^{-1} \]

To find the SI unit of viscosity (η), we start from the formula given: \[ F = \eta A \frac{dV}{dx} \] Where: - \( F \) is the frictional force, - \( \eta \) is the coefficient of viscosity, - \( A \) is the area, ...
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • STATES OF MATTER : GASES AND LIQUIDS

    DINESH PUBLICATION|Exercise Select the correct answer|34 Videos
  • STATES OF MATTER : GASES AND LIQUIDS

    DINESH PUBLICATION|Exercise Comphension 1|5 Videos
  • STATES OF MATTER : GASES AND LIQUIDS

    DINESH PUBLICATION|Exercise Very Short Answer Questions|25 Videos
  • STATES OF MATTER (SOLID STATE CHEMISTRY)

    DINESH PUBLICATION|Exercise ULTIMATE PREPARATORY PACKAGE|21 Videos
  • STRUCTURE OF ATOM

    DINESH PUBLICATION|Exercise Reason Type Questions|1 Videos

Similar Questions

Explore conceptually related problems

The velocity distribution for the flow of a Newtonian fluid between two wide, parallel plates is given by the equation u=(3V)/2[1-(y/h)^(2)] where V is the mean velocity. The fluid has coefficient of viscosity eta Answer the following 3 questions for this situation.

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 :

Knowledge Check

  • The coefficient of viscosity n of a fluid moving steadily between two surfaces is given by the formula, f = nA(dV)/(dx) where f is the frictional force on the fluid, A is an area in the fluid and dV//dx is the velocity gradient inside the fluid at that area. The SI unit of viscosity coefficient n is given as.

    A
    `kgm^(-1)s^(-1)`
    B
    `Nm^(-2)s^(-2)`
    C
    Nil, it is dimensionless constant
    D
    newtons
  • 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 :

    A
    ` [M L ^(-1 ) T^(-1) ] `
    B
    ` [ MLT^(-1) ] `
    C
    `[ML^(-2) T^(-2) ] `
    D
    `[ML ^(-1) T^(-2) ] `
  • 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 :

    A
    ` 10 ^(-4) Nm^(-2) `
    B
    ` 2 xx 10 ^(-4) Nm^ (-2) `
    C
    ` 4 xx 10 ^(-4) Nm^(-2)`
    D
    ` 5 xx 10 ^(-4) Nm^(-2) `.
  • Similar Questions

    Explore conceptually related problems

    Viscous force is represented as F=etaA(dv)/(dx). Where, eta is proportionaly constant called as

    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:

    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))/(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 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 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 , 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 .