Question

A capillary tube of radius $r$ is immersed in water and water rises in it to a height $h$. The mass of water in the capillary tube is $5g$. Another capillary tube of radius $2r$ is immersed in water. The mass of water that will rise in this tube is:

• A. $2.5g$
• B. $5.0g$
• C. $10g$
• D. $20g$

Answer 1

Answer: (C)

$10g$

Given : $r_1=r$ $h_1=h$ $r_2=2r$

Using $rh=constant$

$\therefore$ $r_2{h}_2=r_1{h}_1$

$\left(2r\right)h_2=rh$ $⟹h_2=\frac{h}{2}$

Mass of the water in the capillary $M=\rho \left(\pi r^{2}h\right)$ where $\rho$ is the density of the water

Given : Mass of water that rises in the first capillary $M_1=5$ g

$\therefore$ $5=\rho \pi r^{2}h$ ............(1)

Mass of water that will rise in second capillary $M_2=\rho \left(\pi r_2^2{h}_2\right)$

$M_2=\rho \pi (2r{)}^2\frac{h}{2}=2\times \rho \pi r^{2}h$

$⟹$ $M_2=2\times 5=10$ g