Sangeetha Pulapaka
0

If R is the radius and H is the height

mass of water in first tube = m = volume × density = \pi R^2 Hρ

surface tension = T = \frac{HρgR}{2}  ................................. (1)

Let H' is the height to which water rises in second tube & R' is the radius.

As R' = 2R ----------- given

Hence mass of water in 2nd tube,

M' =   \pi R'^{2} H'ρ

and

surface Tension = \frac{H'ρgR}{2}............................................(2)

surface tension will remain same hence from (1) & (2)

\frac{HρgR}{2} = \frac{H'ρgR}{2}

∴ HR = H'R'

∴ HR = H' × 2R

∴ H = 2H' & H' = (H/2)

∴ mass of water in 2nd tube = M' = \pi R'^{2} H'ρ

= \pi * (2R)^2 * \frac{H}{2} * ρ

= 2 \pi R^2Hρ

M'  = 2M

Sangeetha Pulapaka
0
The mass is doubled
Krishna
0

Capillary action:

If you stick the open end of a capillary tube into water, the water will draw itself right up into the tube.

Mass of the water in the capillary tube M = \pi R^2 H ρ

Where R = radius

H = height

ρ  = Density of liquid

From this we can write  M \propto R (since  ρ, \pi , and H assume as a constants ).

If R increases M also increases.