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 ................
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(d)
Rate of flow $\propto$ pressure difference $\times(\text { radius })^4$
$Q \propto P \times a^4$ $\left\{\because Q=\frac{\pi P r^4}{8 \eta L}\right\}$
So, $\frac{Q_1}{Q_2}=\frac{P_1 a_1^4}{P_2 a_2^4}$
$\frac{Q_1}{Q_2}=\frac{P \times a^4}{4 P \times\left(\frac{a}{4}\right)^4}=\frac{64}{1}$
$\therefore Q_2=\frac{Q_1}{64}=\frac{Q}{64}$
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