First slide

Viscosity

Question

When water flows at a rate Q through a tube of radius r placed horizontally, a pressure difference p develops across the ends of the tube. If the radius of the tube is doubled and the rate of flow halved, the pressure difference will be:

Easy

A

8 P
B

P
C

P / 8
D

P / 32

Solution

Poiseuille’s formula gives the quantity of liquid flowing through a capillary, $Q = \frac{π}{8} \frac{p r^{4}}{η l}$
$i . e . p = \frac{8}{π} Q . \frac{η l}{r^{4}}$
$i f Q' = \frac{Q}{2}, r' = 2 r$
$p' = \frac{8}{π} \frac{Q}{2} . \frac{η l}{{(2 r)}^{4}} = \frac{8}{π} \frac{Q . η l}{r^{4}} \times \frac{1}{32}$
i.e., pressure $p' = \frac{p}{32}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App