Question

A capillary tube of radius $r$ is immersed vertically in a liquid such that liquid rises in it to height $h$ (less than the length of the tube). Mass of liquid in the capillary tube is $m$. If radius of the capillary tube is increased by $50$%, then mass of 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 $\frac{3}{2}m$

$h=\frac{2T\cos \theta }{r\rho g}\Rightarrow h\propto \frac{1}{r}\Rightarrow \frac{h_2}{h_1}=\frac{r_1}{r_2}=\frac{2}{3}$
$(\because r_1=r,r_2=r+50$% of $r=\frac{3}{2}r)$
New mass $m_2=\pi r_2^2{h}_2\rho =\pi {\left(\frac{3}{2}{r}_1\right)}^2\left(\frac{2}{3}{h}_1\right)\rho$
$=\frac{3}{2}\left(\pi r_1^2{h}_1\right)\rho =\frac{3}{2}m$