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 2^{nd} 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 2^{nd} tube = M' = \pi R'^{2} H'ρ

= \pi * (2R)^2 * \frac{H}{2} * ρ

= 2 \pi R^2Hρ

M' = 2M