A U-tube is made up of two capillaries of bores 1.2 m and 2.4 m respectively. The tube is held vertical and partially filled with liquid of surface tension 0.006 N/m and zero contact angle. If estimated difference in level of two menisci is 15 mm, determine mass density of liquid.

SOLUTION:

Given,

Diameter of smaller tube,`d_1=0.0012 m`

Diameter of larger tube,`d_2=0.0024 m`

Let `h_1` and `h_2` be the height of the liquid columns in the two limbs of bore `d_1` and `d_2` respectively. 

Given,`h_1-h_2=0.015`

Now,we have,

`h_1=(4sigma*cos theta)/(gamma*d_1)`

or, `h_1=(4sigma)/(gamma*d_1)`

Similarly,

`h_2=(4sigma)/(gamma*d_2)`

Now,

`h_1-h_2=(4sigma)/(gamma*d_1)-(4sigma)/(gamma*d_2)`

or,`h_1-h_2=(4sigma)/(gamma)*(1/d_1-1/d_2)`

or`0.015=(4**0.06)/(rho**9.81)*(1/0.0012-1/0.0024)`

Thus,`rho=679.58`kg/`m^3`

SOLUTION:

Given,

Diameter of smaller tube,`d_1=0.0012 m`

Diameter of larger tube,`d_2=0.0024 m`

Let `h_1` and `h_2` be the height of the liquid columns in the two limbs of bore `d_1` and `d_2` respectively. 

Given,`h_1-h_2=0.015`

Now,we have,

`h_1=(4sigma*cos theta)/(gamma*d_1)`

or, `h_1=(4sigma)/(gamma*d_1)`

Similarly,

`h_2=(4sigma)/(gamma*d_2)`

Now,

`h_1-h_2=(4sigma)/(gamma*d_1)-(4sigma)/(gamma*d_2)`

or,`h_1-h_2=(4sigma)/(gamma)*(1/d_1-1/d_2)`

or`0.015=(4**0.06)/(rho**9.81)*(1/0.0012-1/0.0024)`

Thus,`rho=679.58`kg/`m^3`