Question

Water flows in a stream line manner through a capillary tube of radius a. The pressure difference being $P$ and the rate of flow is $Q$. If the radius is reduced to $\frac{a}{4}$ and the pressure is increased to $4 P$, then the rate of flow becomes

• A. $4 Q$
• B. $\frac{Q}{2}$
• C. $Q$
• D. $\frac{Q}{64}$

Answer 1

Answer: (D)

Rate of flow $\propto$ pressure difference $\times$ (radius) $^4$
$\left\{\because Q=\frac{\pi P r^{4}}{8 \eta L}\right\}$
$ Q \propto P \times a^{4}$
So, $\frac{Q_{1}}{Q_{2}}=\frac{P_{1} a_{1}^{4}}{P_{2} a_{2}^{4}}$
$\frac{Q_{1}}{Q_{2}}=\frac{P \times a^{4}}{4 P \times\left(\frac{a}{4}\right)^{4}}=\frac{64}{1}$
$\therefore Q_{2}=\frac{Q_{1}}{64}=\frac{Q}{64}$