Properties Of Solids And Liquids Question 237

Question: Water flows in a streamlined manner through a capillary tube of radius a, the pressure difference being P and the rate of flow Q. If the radius is reduced to a/2 and the pressure increased to 2P, the rate of flow becomes

Options:

A) 9.4 m

B) 4.9 m

C) 0.49 m

D) 0.94 m

Show Answer

Answer:

Correct Answer: D

Solution:

Given, $ l _{1}=l _{2}=1, $ and $ \frac{r _{1}}{r _{2}}=\frac{1}{2} $

$ V=\frac{\pi P _{1}r _{1}^{4}}{8\eta l}=\frac{\pi P _{2}r _{2}^{4}}{8\eta l} $

therefore $ \frac{P _{1}}{P _{2}}={{\left( \frac{r _{2}}{r _{1}} \right)}^{4}}=16 $

therefore $ P _{1}=16P _{2} $

Since both tubes are connectedin series, hence pressure difference across combination, $ P=P _{1}+P _{2} $

therefore 1 = $ P _{1}+\frac{P _{1}}{16} $

therefore $ P _{1}=\frac{16}{17}=0.94m $



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