Question

A capillary tube is kept vertical with the lower end in water. If the length of the capillary tube is made $2cm$, then the angle made by water surface in the capillary tube with the wall is:

• A. $0^o$
• B. ${30}^o$
• C. ${60}^o$
• D. ${90}^o$

Answer 1

Answer: (C)

${60}^o$

As we know that capillary rise and fall is given as by following relation,

$h=\frac{4\sigma \cos \theta }{\omega d}$

where, $h=$rise in capillarity

$\omega =$specific weight

$d=$diameter

So, $h\propto 4cos\theta$

$\cos \theta =\frac{2}{4}=0.5$

$\theta ={60}^o$