Question

If a capillary tube of radius \[r\] is dipped in water, the height of water that rises in it is \[h\] and its mass is \[M\]. If the radius of the capillary tube is doubled the mass of water that rises in the tube will be:
(A) \[4M\]
(B) \[2M\]
(C ) \[M\]
(D) \[\dfrac{M}{2}\]

Answer 1

Hint:At first the formula that related the height of the capillary tube with the radius must be noted. Then it must be checked, how the mass varies with radius. Thus, the change in mass can be obtained, on doubling the radius.

Formula Used:
1. \[H = \dfrac{{2T\cos \theta }}{{\rho gr}}\]
\[H = \] Height of the capillary tube
\[T = \] Tension
\[\theta = \]Angle of contact.
\[\rho = \]Density of the liquid
\[g = \] Acceleration due to gravity
\[r = \] Radius of the tube.

2. \[M = \rho \pi {r^2}H\]
Where:
\[H = \] Height of the capillary tube
\[r = \] Radius of the tube.
\[\rho = \]Density of the liquid

Complete Step By Step Solution:
In case of Capillary tube, we know the formula:

\[H = \dfrac{{2T\cos \theta }}{{\rho gr}}\]

Where,
\[H = \] Height of the capillary tube
\[T = \] Tension
\[\theta = \]Angle of contact.
\[\rho = \]Density of the liquid
\[g = \] Acceleration due to gravity
\[r = \] Radius of the tube.
It is clear that if all other factors are considered to be constant, then height of the capillary tube is inversely proportional to the radius of the tube.

\[H \propto \dfrac{1}{r}\]

So, when radius is increased, height of the tube must be decreased linearly.
Therefore, when, radius is halved, then height is doubled.
Further using the formula that relates mass to radius and height.
We know:

\[M = \rho \pi {r^2}H\]

Where:
\[H = \] Height of the capillary tube
\[r = \] Radius of the tube.
\[\rho = \]Density of the liquid
Now, putting the new values of r and H in the above equation:

\[M = \rho \pi {(2r)^2}(\dfrac{H}{2})\]

On solving we get:

\[{M'} = 2\rho \pi {r^2}H\]
Thus,

\[{M'} = 2M\]

So, option (B) is correct.

Note:
The maximum possible height that the tube may attain is \[H = \dfrac{{2T}}{{\rho gr}}\]
This happens when\[\theta = 0\].
Forces between the tubes are adhesive forces that are between the liquid molecules and that of the walls of the tube.