The viscous resistance of a tube to liquid flow is $R$ . Its resistance for a narrow tube of same length and $\frac{1}{3}$ times radius is :

\frac{R}{3}

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3 R

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27 R

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81 R

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Solution

The correct option is C $81 R$
From Poiseulle's eqn,,
$Q = \frac{\frac{P}{8 n L}}{π r^{4}}$
So,
$R = \frac{8 η L}{π r^{4}}$ ............... (1)
New R for the narrow tube
$= (\frac{8 η L}{π r^{4}}) \times 3^{4} = 81 R$

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A wire of radius r has resistance R . If it is stretched to a radius of $\frac{3 r}{4}$ , its resistance becomes

{16 c m}^{3}

R

\frac{L}{3}

\frac{R}{2}

P

A v i s c o u s l i q u i d ρ = 1.2 \times 10^{3} k g / m^{3}, μ = 0.0048 P a . s f l o w s a t a r a t e o f 10 m l / s t h r o u g h a h o r i z o n t a l, 4 m m d i a m e t e r t u b e . T h e p r e s s u r e d r o p a l o n g a 1 m l e n g t h o f t h e t u b e w h i c h i s f a r f r o m t h e t u b e e n t r a n c e s o t h a t t h e f l o w i s f u l l y d e v e l o p e d i s

k P a

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