Question

Two liquids of densities d1, and d2, and coefficient of
viscosities ni and n2 are found to flow through a capillary
tube at the same rate. Find the ratio of n, and n2?​

Answer 1

Answer:

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Answer 2

Answer:

$\sf \dfrac{\eta_{1}}{\eta_{2}} = \dfrac{\rho_1}{\rho_2}$

Explanation:

Given :

Two liquid of density $\sf d_1$ and $\sf d_2$ of coefficient of  viscosity $\sf \eta_1$ and $\sf \eta_2$ .

Flowing through a capillary tube at same rate.

We are asked to find ratio of $\sf n_1$ and $\sf n_2$ .

We have Poiseuille's formula :

$\sf V=\dfrac{\pi}{8}.\dfrac{P \ r^4}{\eta \ \ell}$

But we know :

$\sf P=\rho \ h \ g$

$\sf V=\dfrac{\pi}{8}.\dfrac{P \ r^4}{\eta \ \ell} =\dfrac{\pi \ h \ \rho \ g \ r^4}{8 \ \eta \ \ell}$

$\sf \longrightarrow \ \dfrac{\rho}{\eta} =constant$

Therefore :

$\sf \dfrac{\eta_{1}}{\eta_{2}} = \dfrac{\rho_1}{\rho_2}$

Hence we get required answer.