Question

A capillary tube of radius ${}^{\prime }{r}^{\prime }$ is immersed in water and water rises in to a height ${}^{\prime }{h}^{\prime }$. Mass of water in the capillary tube is $5\times {10}^{-3}kg$. Another capillary tube of radius $\left(\frac{r}{2}\right)$ is immersed in water. The mass of water that will rise in this tube is:

• A. $7.5\times {10}^{-3}kg$
• B. $1\times {10}^{-3}kg$
• C. $2.5\times {10}^{-3}kg$
• D. $5\times {10}^{-3}kg$

Answer 1

Answer: (C)

$2.5\times {10}^{-3}kg$

By balancing forces on the rising water,

$F=Scos\theta \times l=mg$

$S\left(2\pi r\right)=\rho \left(\pi r^{2}h\right)g$

$h=\frac{2Scos\theta }{\rho gr}$

Also, mass is $m=\rho \left(\pi r^{2}h\right)g=\rho \pi g{r}^2\left(\frac{2Scos\theta }{\rho gr}\right)=2\pi Scos\theta r$

Since mass is proportional to the radius, hence as radius is halves, mass of rising water will also get halves. Hence $m=2.5\times {10}^{-3}kg$