The rate of flow of liquid in a tube of radius $r$, length $l$, whose ends are maintained at a pressure difference $p$ is $V=\frac{\pi N p r^4}{\eta l}$, where $\eta$ is coefficient of the viscosity and $N$ is

JEE PHYSICS MCQ: MECHANICAL PROPERTIES OF FLUIDS

1 Medium

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Remembering +4 | -1

The rate of flow of liquid in a tube of radius

r

, length

l

, whose ends are maintained at a pressure difference

p

V=\frac{\pi N p r^4}{\eta l}

, where

\eta

is coefficient of the viscosity and

N

View answer and explanation

Correct option: B

\frac{1}{8}

Option B (

\frac{{1}}{{8}}

) is correct because Poiseuille's equation for the rate of flow of liquid through a tube is

V = \frac{{\pi p r^4}}{{8 \eta l}}

, so N = 1/8.

Practice All Chapter MCQs →

80 \mathrm{~kg}

20 \mathrm{~cm}

0.5 \mathrm{~cm}

0.1 \mathrm{~cm}

15 \mathrm{~m} / \mathrm{s}

5 \mathrm{~m}

=10^{-2}

m_1

d_1

d_2

m_2