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Question: The rate of steady volume flow of water through a capillary tube of length 'l' and radius 'r' under ...

The rate of steady volume flow of water through a capillary tube of length 'l' and radius 'r' under a pressure difference of P is V. This tube is connected with another tube of the same length but half the radius in series. Then the rate of steady volume flow through them is (The pressure difference across the combination is P)

A

(a) racV16 rac{V}{16}

A

(b) racV17 rac{V}{17}

A

(c) rac16V17 rac{16V}{17}

A

(d) rac17V16 rac{17V}{16}

Explanation

Solution

(b)

Rate of flow of liquid V=PRV = \frac{P}{R}

where liquid resistance R=8ηlπr4R = \frac{8\eta l}{\pi r^{4}}

For another tube liquid resistance

R=8ηlπ(r2)4=8ηlπr4.16=16RR' = \frac{8\eta l}{\pi\left( \frac{r}{2} \right)^{4}} = \frac{8\eta l}{\pi r^{4}}.16 = 16R

For the series combination

VNew=PR+R=PR+16R=P17RV_{New} = \frac{P}{R + R'} = \frac{P}{R + 16R} = \frac{P}{17R}

=V17= \frac{V}{17}.