Q. 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
A) $8\ p$
B) $p$
C) $p/8$
D) $p/32$
Explanation
(d) Poiseuille’s formula gives the quantity of liquid flowing
Through a capillary, $Q = \frac{\pi}{8}\frac{pr^{4}}{\eta l} \Rightarrow p = \frac{8}{\pi}Q.\frac{\eta l}{r^{4}}$
If $Q^{'} = \frac{Q}{2},r^{'} = 2r$
$\frac{p^{'}}{n} = \frac{8}{\pi}\frac{Q}{2}\frac{\eta l}{(2r)^{4}} = \frac{8}{\pi}\frac{Q.\eta l}{r^{4}} \times \frac{1}{32}$
$i.e.,$ pressure $p^{'} = \frac{p}{32}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.