Question

60. The mass of water that rises in capillary tube of radius R is M. The mass of water that rises in tube of radius 2R (a) M(b) M/2
(C) 2M
(d) 4M​

Answer 1

If R is the radius and H is the height

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

2

Hρ

surface tension = T = \frac{HρgR}{2}

2

HρgR

................................. (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'πR

′2

H

′

ρ

and

surface Tension = \frac{H'ρgR}{2}

2

H

′

ρgR

............................................(2)

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

\frac{HρgR}{2} = \frac{H'ρgR}{2}

2

HρgR

=

2

H

′

ρgR

∴ HR = H'R'

∴ HR = H' × 2R

∴ H = 2H' & H' = (H/2)

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

′2

H

′

ρ

= \pi * (2R)^2 * \frac{H}{2} * ρπ∗(2R)

2

∗

2

H

∗ρ

= 2 \pi R^2Hρ2πR

2

Hρ

M' = 2M