Question

Two different vertical positions (a) and (b) of a capillary tube are shown in the figure with the lower end inside water. For position (a) Contact angle is ${45}^o$ and water rises to height h above the surface of water, while for position (b) height of the tube outside water is kept insufficient and equal to $\frac{h}{\sqrt{2}}$. Then contact angle becomes:

• A. ${30}^o$
• B. ${45}^o$
• C. ${60}^o$
• D. ${\cos }^{-1}\left(\frac{1}{2\sqrt{2}}\right)$

Answer 1

The correct option is C ${60}^o$

The height of the liquid which rises inside a capillary is:

$h=\frac{2T\cos \theta }{r\rho g}$

In case(a):

$h=\frac{2T\cos {45}^0}{r\rho g}$ ....(1)

In case (b):

$\frac{h}{\sqrt{2}}=\frac{2T\cos \theta }{r\rho g}$ ....(2)

Dividing (2) by (1) we get:

$\cos \theta =\frac{1}{2}=\cos {60}^0$

$\therefore \theta ={60}^0$