Question

Two liquids of densities $D_1$ and $D_2$ are flowing in identical capillary tubes under the same pressure difference. If $T_1$ and $T_2$ are the time taken for the flow of equal quantities (Mass) of liquids. What will be the ratio of the coefficient of viscosity of liquids?

Answer 1

Step 1: Given parameters

Two liquids have different densities $D_1$ and $D_2$ at time $T_1$ and $T_2$ respectively.

Step 2: Formula used

According to Poiseuille’s equation, the volume($V$) of liquid flowing per second (say $T$) in a capillary tube, is given by

$Q=\frac{V}{T}=\frac{\pi r^4∆P}{8\eta L}......\left(1\right)$.

Where, $T$ is the time, $L$is the length of the pipe, $\eta$ is the dynamic viscosity, $∆P$ is the pressure difference between the two ends, and, $r$ is the radius of the pipe

Now mass ($m$) is and density ($D$) related to volume as $V=\frac{m}{D}$, so equation (1) can be written as.

$Q_m=\frac{m}{T}=\frac{\pi r^4∆P}{8\eta L}D$

Now for each liquid,

$$\begin{gathered} \frac{m_1}{T_1}=\frac{\pi r^4∆P}{8{\eta }_{1}L}{D}_1.......\left(2\right) \\ \frac{m_2}{T_2}=\frac{\pi r^4∆P}{8{\eta }_{2}L}{D}_2........\left(3\right) \end{gathered}$$

Step 3: Calculate the ratio of the coefficient of viscosity of liquids

Using $m_1=m_2$ and dividing (2) by (3), we get

$$\begin{gathered} \Rightarrow \frac{T_2}{T_1}=\frac{D_1}{D_2}\times \frac{{\eta }_2}{{\eta }_1} \\ \Rightarrow \frac{{\eta }_1}{{\eta }_2}=\frac{D_1{T}_1}{D_2{T}_2} \end{gathered}$$

Hence, the ratio ${\eta }_1:{\eta }_2=D_1{T}_1:D_2{T}_2$.