Question

When a clean lengthy capillary tube is dipped vertically in a beaker containing water, the water rises to a height of $8$ cm. What will happen if another capillary tube of length $4$ cm and the same radius is dipped vertically into the same beaker containing water ?

• A. Water will flow out like a fountain.
• B. Water will rise to a height of $4$ cm only and the angle of contact will be zero.
• C. Water will rise to a height of $4$ cm only and the angle of contact will be 60${}^o$
• D. Water will not rise at all.

Answer 1

The correct option is C Water will rise to a height of $4$ cm only and the angle of contact will be 60${}^o$

We may think that water will flow out of the capillary tube. But the water will try to change it's contact angle to adjust the pressure differences.

Let $\Delta P$ be pressure difference between risen water and atmosphere in the capillary tube.

Initially $\Delta P$=$\frac{2T}{R}=\rho g\left(8\right)$ because contact angle is $0^0$

Now let us assume water will have contact angle ${\theta }^0$

$\Delta P=\frac{2Tcos\theta }{R}=\rho g\left(4\right)$

From the above two equations we get $cos\left(\theta \right)=\frac{1}{2}$ which implies $\theta ={60}^0$

The water rises to $4cm$ with contact angle ${60}^0$