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