Question

Water flows in a stream line manner through a capillary tube of radius $a,$ the pressure difference being $P$ and the rate flow $Q.$ If the radius is reduced to $\frac{a}{2}$ and the pressure is increased to $2P,$ the rate of flow becomes :

• A. $4Q$
• B. $Q$
• C. $\frac{Q}{2}$
• D. $\frac{Q}{8}$

Answer 1

The correct option is D $\frac{Q}{8}$

$Q^{\prime }=\frac{\pi \left(2P\right){\left(\frac{a}{2}\right)}^4}{8\eta l}=\frac{Q}{8}$

$[\because Q=\frac{\pi P{a}^4}{8\eta l}]$