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 $5\text{g}$. Another capillary tube of radius $2r$ is immersed in water. The mass of water that will rise in this tube is:

• A. $2.5\text{g}$
• B. $5.0\text{g}$
• C. $10\text{g}$
• D. $20\text{g}$

Answer 1

The correct option is C $10\text{g}$

Let $h$ be the height through which the liquid rise in the capillary tube of radius $r$.
$\therefore h=\frac{2T\cos \theta }{r\rho g}$
where
$T=$ surface tension
$\theta =$ contact angle
$\rho =$ density of liquid

We know
Mass = volume $\times$ density
$\therefore$ Mass of liquid in first tube
$m=\left(\pi r^{2}h\right)\times \rho$
$=\pi r^2\times \frac{2T\cos \theta }{r\rho g}\times \rho$
$=\frac{\pi r2T\cos \theta }{g}$
$\therefore m\propto r$
$\Rightarrow$ Mass of liquid in 2nd capillary tube
$m^{\prime }\propto \left(2r\right)$
$\therefore \frac{m^{\prime }}{m}=\frac{2r}{r}=2$
$\Rightarrow m^{\prime }=2\times 5\text{g}=10\text{g}$