Properties Of Solids And Liquids Question 226
Question: We have two (narrow) capillary tubes T1 and T2. Their lengths are l1 and l2 and radii of cross-section are r1 and r2 respectively. The rate of flow of water under a pressure difference P through tube T1 is 8cm3/sec. If l1 = 2l2 and r1 =r2, what will be the rate of flow when the two tubes are connected in series and pressure difference across the combination is same as before (= P)
Options:
A)4 cm3/sec
B)(16/3) cm3/sec
C)(8/17) cm3/sec
D)None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ V=\frac{\pi {{\Pr }^{4}}}{8\eta l}=\frac{8cm^{3}}{\sec } $
For composite tube $ V _{1}=\frac{P\pi r^{4}}{8\eta \left( l+\frac{l}{2} \right)}=\frac{2}{3}\frac{\pi Pr^{4}}{8\eta l} $
$ =\frac{2}{3}\times 8=\frac{16}{3}\frac{cm^{3}}{\sec } $
$ \left[ \because \ \ l _{1}=l=2l _{2}\ or \ l _{2}=\frac{l}{2} \right] $
