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 connected in series, hence pressure difference across combination, $ P=P _{1}+P _{2} $
therefore 1 = $ P _{1}+\frac{P _{1}}{15} $
therefore $ P _{1}=\frac{16}{17}=0.94 $
BETA
