Question

If a capillary tube of radius r is immersed in a liquid, then the liquid rises to a height h. The corresponding mass of liquid column is m. The mass of water that would rise in another capillary tube of twice the radius is:

• A. $2m$
• B. $5m$
• C. $3m$
• D. $4$
• E. $m/2$

Answer 1

The correct option is A $2m$

Given, radius of first capillary tube$=$r
Height of rised liquid in first capillary tube$=$h
Mass of liquid in first capillary tube$=$m
We know that,
$m\propto \pi r^{2}h$
or $m\propto r^{2}h$
rh$=$constant
So, $h\propto \frac{1}{r}\Rightarrow m\propto r$
Now, $\frac{m_2}{m_1}=\frac{r_2}{r_1}$
$\Rightarrow \frac{m_2}{m_1}=\frac{r_2}{r_1}=2$
$m_2=2m_1$ or $m_2=2$m
$\{\because m_1=m\}$.