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Viscosity

Two liquids o...

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

\frac{d_{1} t_{1}}{d_{2} t_{2}}

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B

\frac{t_{1}}{t_{2}}

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C

\frac{d_{2} t_{2}}{d_{1} t_{1}}

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D

\sqrt{\frac{d_{1} t_{1}}{d_{2} t_{2}}}

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Solution

The correct option is C $\frac{d_{1} t_{1}}{d_{2} t_{2}}$
According to poiseuille's equation, Volume of liquid flowing per second in a capillary tube is given by
$Q = \frac{V}{t} = \frac{π r^{4} Δ P}{8 η L}$
So mass of liquid flowing in a capillary tube is given by
$Q_{m} = \frac{m}{t} = \frac{π r^{4} Δ P}{8 η L} d$
$\Rightarrow \frac{m_{1}}{t_{1}} = \frac{π r^{4} Δ P}{8 η_{1} L} d_{1} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (I)$
$\Rightarrow \frac{m_{2}}{t_{2}} = \frac{π r^{4} Δ P}{8 η_{2} L} d_{2} . . . . . . . . . . . . . . . . . . . . . (I I)$
Dividing $(I)$ by $(I I)$ , we have
$\frac{t_{2}}{t_{1}} = \frac{d_{1}}{d_{2}} \times \frac{η_{2}}{η_{1}} ∵ m_{1} = m_{2}$
$\Rightarrow \frac{η_{1}}{η_{2}} = \frac{d_{1} t_{1}}{d_{2} t_{2}}$

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