Question

A capillary tube of radius $r$ is immersed vertically in a liquid such that the liquid rises in it to height $h$ (less than the length of the tube). Mass of the liquid in the capillary tube is $m$. If the radius of the capillary tube is increased by $50\%$, then the mass of the liquid that will rise in the tube is:

• A. $\frac{2}{3}m$
• B. $m$
• C. $\frac{3}{2}m$
  
• D. $\frac{9}{4}m$
  

Answer 1

The correct option is C $\frac{3}{2}m$


From the formula for capillary rise,
$h=\frac{2T\cos \theta }{r\rho g}$
$\Rightarrow h\propto \frac{1}{r}$
[$T\&\cos \theta$ are same for same liquid and tube]
Applying for two different radii of same tube, the term $\frac{2T\cos \theta }{\rho g}=\text{constant}$
$\Rightarrow \frac{h_2}{h_1}=\frac{r_1}{r_2}=\frac{2}{3}$
$(\because r_1=r\&r_2=r+(50\%\text{of}r)=\frac{3}{2}r)$
$\Rightarrow h_2=\frac{2}{3}{h}_1$
Initial mass of the liquid column in the capillary tube is given by,
$m=V\times \rho =(\pi r^2\times h_1)\times \rho$
$\Rightarrow m=(\pi r^2\times h_1)\times \rho$
Final mass of the liquid column in the capillary tube is,
$m_2=(\pi r_2^2\times h_2)\times \rho =\pi {\left(\frac{3}{2}r\right)}^2\left(\frac{2}{3}{h}_1\right)\rho$
$m_2=\frac{3}{2}\left(\pi r^2{h}_1\right)\rho$
$\therefore m_2=\frac{3}{2}m$