Question

If a $10\text{cm}$ long capillary tube with $0.2\text{mm}$ internal diameter open at both ends is slightly dipped in water having surface tension $80\text{dyne/cm}$, then:
[$g=981{\text{g/cm}}^2$]

• A. Height to which water level will rise will be $16.31\text{cm}$.
• B. Height to which water level will rise will be equal to length of capillary.
• C. Radius of curvature of meniscus will be $0.163\text{mm}$.
• D. Radius of curvature of meniscus will be $0.0163\text{mm}$.

Answer 1

Answer: (B), (C)

Radius of curvature of meniscus will be $0.163\text{mm}$.

Given that,
Surface tension of water $\left(T\right)=80\text{dyne/cm}$
Radius of tube $\left(r\right)=\frac{0.2}{2}=0.1\text{mm}=0.01\text{cm}$
Density of water $\left(\rho \right)=1{\text{g/cm}}^3$
Angle of contact $\left(\theta \right)=0^{\circ }$
Let $h$ be the height to which water rises in the capillary.
$h=\frac{2T\cos \theta }{r\rho g}=\frac{2\times 80\times \text{cos}\theta }{0.01\times 1\times 981}=16.31\text{cm}$
So, $h=16.31\text{cm}$
But length of capillary tube $\left(L\right)=10\text{cm}$
Because tube is of insufficient length, the water will rise upto the upper end of the tube.
So, $h^{\prime }=10\text{cm}$ is the capillary rise.

Since radius of curvature of meniscus $R=\frac{r}{\cos \theta }$, hence
$h=\frac{2T}{R\rho g}$. Here, $T,\rho \&g$ are constant.
$\Rightarrow hR=$ constant
i.e Liquid meniscus will adjust its radius of curvative $R$ in such way that
$R^{\prime }{h}^{\prime }=Rh$
$\Rightarrow R^{\prime }=\frac{Rh}{h^{\prime }}=\frac{r}{\cos \theta }\times \frac{h}{h^{\prime }}=\frac{0.01\times 16.31}{10}$
$\Rightarrow R^{\prime }=0.01631\text{cm}=0.163\text{mm}$
Therefore, options (b) and (c) are correct.