SATHEE: Properties Of Solids And Liquids Question 226

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:

$V = \frac{π {Pr}^{4}}{8 η l} = \frac{8 c m^{3}}{\sec}$

For composite tube $V_{1} = \frac{P π r^{4}}{8 η (l + \frac{l}{2})} = \frac{2}{3} \frac{π P r^{4}}{8 η l}$

$= \frac{2}{3} \times 8 = \frac{16}{3} \frac{c m^{3}}{\sec}$

$[∵ l_{1} = l = 2 l_{2} o r l_{2} = \frac{l}{2}]$

Play Store

App Store

Ask SATHEE

Welcome to SATHEE !
Select from 'Menu' to explore our services, or ask SATHEE to get started. Let's embark on this journey of growth together! 🌐📚🚀🎓

I'm relatively new and can sometimes make mistakes.
If you notice any error, such as an incorrect solution, please use the thumbs down icon to aid my learning.
To begin your journey now, click on "I understand".

Properties Of Solids And Liquids Question 226

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu