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Byju's Answer

Standard XII

Physics

Tube of Insufficient Length

Two liquids a...

Question

Two liquids are allowed to flow through two capillary tubes of lengths in the ratio $1 : 2$ and radii in the ratio $2 : 3$ under the same pressure difference. If the volume rates of flow of the liquids are in the ratio $8 : 9$ , the ratio of their coefficients of viscosity is :

1 : 3

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3 : 1

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4 : 9

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9 : 4

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Solution

The correct option is C $4 : 9$

From poiseullies eqn, $Q = \frac{π p r^{4}}{8 η L}$
According to Question, $\frac{8}{9} = \frac{2^{4} \times η_{2}}{3^{4} \times η_{1}} 2$
$η_{1} : η_{2} = \frac{9 \times 2^{2}}{3^{2} \times 8} = \frac{4}{9}$

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Two liquids are allowed to flow through two capillary tubes of lengths in the ratio 1:2 and radii in the ratio 2:3 under the same pressure difference. If the volume rates of flow of the liquids are in the ratio 8:9, the ratio of their coefficients of viscosity is :

The correct option is C 4:9 From poiseullies eqn,Q=πpr48ηLAccording to Question,89=24×η234×η12η1:η2=9×2232×8=49

Two liquids are allowed to flow through two capillary tubes of lengths in the ratio $1 : 2$ and radii in the ratio $2 : 3$ under the same pressure difference. If the volume rates of flow of the liquids are in the ratio $8 : 9$ , the ratio of their coefficients of viscosity is :

The correct option is C $4 : 9$

From poiseullies eqn, $Q = \frac{π p r^{4}}{8 η L}$
According to Question, $\frac{8}{9} = \frac{2^{4} \times η_{2}}{3^{4} \times η_{1}} 2$
$η_{1} : η_{2} = \frac{9 \times 2^{2}}{3^{2} \times 8} = \frac{4}{9}$