Question

If $'M'$ is the mass of water that rises in a capillary tube of radius $'r'$, then mass of water which will rise in a capillary tube of radius $'2r'$ is :

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

Answer 1

Answer: (C)

Height of liquid rise in capillary tube h = $\frac{2T cos \theta_c}{\rho rg}$
$\Rightarrow $ h \propto $\frac{1}{r}$
when radius becomes double height become half
$\therefore \, h' = \frac{h}{2}$
Now, M = $\pi r^2 h \times \rho$
and M' = $\pi (2r^2) (h/2) \times \rho = 2M$
Option (3)