Question

The factor by which the pressure across a capillary tube is to be changed, so that the rate of flow of a liquid is doubled, when the length and the radius of the tube are doubled is :

• A. $4$
• B. $8$
• C. $1/8$
• D. $1/4$

Answer 1

Answer: (D)

$1/4$

Applying Poiseuilli's eqn,

$Q=\frac{\pi r^{4}P}{8\eta L}$ .............(1)

$\to P=\frac{Q\times 8\eta L}{\pi r^4}$

In 2${}^{nd}$ case,

$P^{\prime }=\frac{2Q\times 8\eta \left(2L\right)}{\pi (2r{)}^4}=P/4$

So, factor by which pressure would change is $\frac{1}{4}$.