Question

Under a pressure head, the rate of orderly volume flow of liquid through a capillary tube is Q, if the length of the capillary is doubled and the diameter of the bore is halved, the rate of flow would become

• A. $\frac{Q}{4}$
• B. $\frac{Q}{16}$
• C. $\frac{Q}{32}$
• D. $\frac{Q}{8}$

Answer 1

The correct option is C $\frac{Q}{32}$

Since
$Q=\frac{\pi p{r}^4}{8\eta l}$
So when diameter is halved i.e.radius is halved and length is doubled,
$Q^{\prime }=\frac{\pi p(\frac{r}{2}{)}^4}{8\eta \times 2l}=Q/32$
$Q$ becomes $Q/32$