Two liquids of densities d_1 and d_2 are flow | vedclass

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(A) According to Poiseuille's equation,the volume of liquid flowing per second $(Q)$ in a capillary tube is given by:$Q = \frac{V}{t} = \frac{\pi r^4

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$\frac{d_1 t_1}{d_2 t_2}$

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(A) According to Poiseuille's equation,the volume of liquid flowing per second $(Q)$ in a capillary tube is given by:$Q = \frac{V}{t} = \frac{\pi r^4 \Delta P}{8 \eta L}$Since mass $m = V \cdot d$,the mass of liquid flowing per second is:$\frac{m}{t} = \frac{\pi r^4 \Delta P}{8 \eta L} \cdot d$For two liquids with equal mass $m$ flowing through identical tubes under the same pressure difference $\Delta P$:$\frac{m}{t_1} = \frac{\pi r^4 \Delta P}{8 \eta_1 L} d_1 \implies \eta_1 = \frac{\pi r^4 \Delta P \cdot d_1 t_1}{8 L m} \dots (I)$$\frac{m}{t_2} = \frac{\pi r^4 \Delta P}{8 \eta_2 L} d_2 \implies \eta_2 = \frac{\pi r^4 \Delta P \cdot d_2 t_2}{8 L m} \dots (II)$Dividing equation $(I)$ by $(II)$:$\frac{\eta_1}{\eta_2} = \frac{d_1 t_1}{d_2 t_2}$

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 times taken for the flow of equal quantities (mass) of liquids,then the ratio of the coefficient of viscosity of the liquids must be:

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