Question

A capillary tube is attached horizontally at a constant head arrangement. If the radius of the capillary tube is increased by 10%, the rate of flow of liquid changes by about:

• A. -40%
• B. +40%
• C. +21%
• D. +46%

Answer 1

The correct option is D +46%

Let the radius be r and preassure difference be p
Q= flow = $\frac{\pi (p_1-p_2)r^4}{\delta \left(\eta \right)\left(\iota \right)}$
$\frac{\pi (p_1-p_2)r^4}{\delta \left(\eta \right)\left(\iota \right)}$
Now,
new $r=\frac{110r}{100}=1.1r$
$Q_{new}$=$\frac{\pi (p_1-p_2)(1.1{)}^4{r}^4}{3\eta \iota }$
so, Increase i rate of flow =$\frac{Q_{new}-Q}{Q}\times 100$%
= $\left(\frac{1.464-1}{1}\right)\times 100$%
=46.4%