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

In the second capillary tube, the water's mass will rise.

rh = constant, as we already know.

$\therefore {\mathrm{r}}_2{\mathrm{h}}_2={\mathrm{r}}_1{\mathrm{h}}_1$

$\left(2\mathrm{r}\right){\mathrm{h}}_2=\mathrm{rh}\Rightarrow {\mathrm{h}}_2=\frac{\mathrm{h}}{2}\mathrm{\_\_\_\_\_\_}\left(1\right)$

Now, the volume of water that ascends in our first capillary tube;

$\mathrm{Density}=\frac{\mathrm{mass}}{\mathrm{volume}}$

${\mathrm{M}}_1=\mathrm{\rho }\left({\mathrm{\pi r}}^2\mathrm{h}\right)$

$5=\mathrm{\rho }\left({\mathrm{\pi r}}^2\mathrm{h}\right)$

${\mathrm{M}}_2=\mathrm{\rho }\left(\mathrm{\pi }{{\mathrm{r}}^2}_2{\mathrm{h}}_2\right)\mathrm{\_\_\_\_\_\_}\left(2\right)$

from equation 1 and 2

${\mathrm{M}}_2=\mathrm{\rho }\left(\mathrm{\pi }{{\mathrm{r}}^2}_2{\mathrm{h}}_2\right)=2\left({\mathrm{\rho \pi r}}^2\frac{\mathrm{h}}{2}\right)=2{\mathrm{M}}_1$

$\Rightarrow {\mathrm{M}}_2=2\times 5=10\mathrm{g}$

Hence the correct answer is 10g.