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 time taken for the flow of equal quantities (mass) of liquids, then the ratio of coefficient of viscosity of liquids must be:
(A) $\dfrac{{{d_1}{t_1}}}{{{d_2}{t_2}}}$
(B) \[\dfrac{{{t_1}}}{{{t_2}}}\]
(C) \[\dfrac{{{d_2}{t_2}}}{{{d_1}{t_1}}}\]
(D) $\sqrt {\dfrac{{{d_1}{t_1}}}{{{d_2}{t_2}}}} $

Answer 1

Hint: The Poiseuille’s law provides information about the rate of the flow of liquid in a capillary tube when there is pressure difference at both the ends of the tube. And the rate of flow gives the amount of the fluid passed in unit times from a cross-section.

Complete step by step answer:
Given,
It is given that both the capillary tubes are in the same pressure difference, let it is $\Delta P$ .
The density of the first liquid is ${d_1}$ , and the density of the second liquid is \[{d_2}\] .
It is given that the time taken by the first and second liquid is ${t_1}$and ${t_2}$ respectively for the flow of the same mass $m$.
The expression for the rate of flow of the liquid across the end of a capillary tube is given as follows,
$Q = \dfrac{{\pi {r^4}\Delta P}}{{8\mu L}}$
Here, $\mu $is the viscosity of the liquid and $r$ is the radius of the capillary tube and $L$ is the length of the tube.
Now, we write the expression for the rate of flow of the liquid for first tube,
${Q_1} = \dfrac{{\pi r_1^4\Delta P}}{{8{\mu _1}{L_1}}}\;...............{\rm{(1)}}$
Now, we write the expression for the rate of flow of the liquid for second tube,
${Q_2} = \dfrac{{\pi r_2^4\Delta P}}{{8{\mu _2}{L_2}}}\;...............{\rm{(2)}}$
It is given that the tubes are identical, so the value of $r$ and $L$ will be the same for both the tube, and we can write,
${r_1} = {r_2}$ and ${L_1} = {L_2}$
Divide the equation (1) by the equation (2).
$
\Rightarrow\dfrac{{{Q_1}}}{{{Q_2}}} = \dfrac{{\dfrac{{\pi r_1^4\Delta P}}{{8{\mu _1}{L_1}}}}}{{\dfrac{{\pi r_2^4\Delta P}}{{8{\mu _2}{L_2}}}}}\\
\Rightarrow\dfrac{{{Q_1}}}{{{Q_2}}} = \dfrac{{{\mu _2}}}{{{\mu _1}}}\;..................{\rm{(3)}}
$
Also, we know that the expression for the discharge is,
$
Q = \dfrac{V}{t}\\
\Rightarrow Q = \dfrac{m}{{d \times t}}
$
Here, $V$ is the volume of the liquid, $m$ is the mass of the liquid, $d$ is the density of the liquid and $t$ is the time.
Now, substitute the value of discharge for both the liquid in the equation (3).
$
\Rightarrow\dfrac{{\dfrac{m}{{{d_1} \times {t_1}}}}}{{\dfrac{m}{{{d_2} \times {t_2}}}}} = \dfrac{{{\mu _2}}}{{{\mu _1}}}\\
\Rightarrow\dfrac{{{\mu _1}}}{{{\mu _2}}} = \dfrac{{\dfrac{m}{{{d_2} \times {t_2}}}}}{{\dfrac{m}{{{d_1} \times {t_1}}}}}\\
\therefore\dfrac{{{\mu _1}}}{{{\mu _2}}} = \dfrac{{{d_1} \times {t_1}}}{{{d_2} \times {t_2}}}
$
Therefore, ratio of the viscosity will be $\dfrac{{{d_1}{t_1}}}{{{d_2}{t_2}}}$ and the correct answer is option (A).

Note: The viscosity of a fluid provides information about the resistance of the fluid against the motion, so the fluid having low viscosity can move quickly on a surface, but the fluid that has a high value of viscosity, moves slower.