Chapter 10, Problem 50P

To determine

The viscosity of the liquid is to be determined on the basis of given information..

Answer to Problem 50P

The viscosity of the liquid is $7.4\times {10}^{-2}Pa\sec$ .

Explanation of Solution

Given info:

Diameters of two concentric cylinders are,

  $\begin{array}{l}{d}_1=10.20cm=0.1020m\Rightarrow r_1=0.051m\\ d_2=10.60cm=0.1060m\Rightarrow r_2=0.053m\end{array}$

  $depth=12cm=0.12m$

Torque, $\tau =0.024N$

The rotational speed of the cylinder, $v=62rev/\min$

Formula used:

The torques are given as,

  $\tau =F\times r$

The viscosity formula is,

  $\eta =\frac{F}{\left(Av/l\right)}$

Calculation:

The tangential force of acting on the inner cylinder is,

  $F=\frac{\tau }{r}=\frac{0.024N}{0.051}=0.471N$

Now, according to the definition of the coefficient of viscosity is

  $\eta =\frac{F}{\left(Av/l\right)}=\frac{0.471\times l}{\left(2\pi rh\right)\left(\omega r\right)}$

Where, $l$ is the thickness of the cylinder, so

  $l=r_2-r_1=0.053m-0.051m=0.002m$

Now use these values in the above equation,

  $\begin{array}{l}\eta =\frac{F}{\left(Av/l\right)}=\frac{0.471\times l}{\left(2\pi rh\right)\left(\omega r\right)}=\frac{0.471\times 0.002}{2\times 3.14\times 0.051\times 12\times \left(2\pi r\frac{62}{60}\right)}=\frac{0.471\times 0.002\times 60}{2\times 3.14\times 0.051\times 12\times \left(2\times 3.14\times 0.051\times 62\right)}\\ =\frac{0.05652}{76.318}=7.4\times {10}^{-2}Pa\sec \end{array}$

Conclusion:

The viscosity is $7.4\times {10}^{-2}Pa\sec$