Question

Water rises in a capillary tube through a height h if the tube is inclined to the liquid surface 45° the liquid will rise in the tube up to its length equal to

Answer 1

Water rises in a capillary tube through a height h if the tube is inclined to the liquid surface 45° the liquid will rise in the tube up to its length equalto.

Answer 2

Answer:$h' = \dfrac{h}{\sqrt2}$

Explanation:

Given that,

Inclined angle $\theta = 45 ^{0}$

Let us consider the contact angle of the liquid with the tube is $45^{0}$ and  the liquid will rise in the tube up its length equal to h'.

The height of capillary tube is

$h = \dfrac{2Tcos\theta}{r\rho g}$

If the contact angle of water with the tube is zero,then the height of capillary is

$h = \dfrac{2Tcos 0^{0}}{r\rho g}$.....(I)

If the contact angle of water with the tube is $45^{0}$,then the height of capillary is

$h' = \dfrac{2Tcos 45^{0}}{r\rho g}$.....(II)

Divided by equation (I) by equation (II)

$\dfrac{h}{h'} = \dfrac{2T}{r\rho g}\times\dfrac{r\rho g\sqrt2}{2T}$

$\dfrac{h}{h'} = \sqrt2$

$h' = \dfrac{h}{\sqrt2}$

Hence, the liquid will rise in the tube up to its length is $\dfrac{h}{\sqrt2}$