The lower end of a capillary tube is at a depth of 12cm and the water rises 3cm in it. The mouth pressure required to blow an air bubble at the lower end will be X cm of water column where X is

The correct answer is 15.

To find the mouth pressure required to blow an air bubble at the lower end of the capillary tube, we need to consider two components of pressure that the applied pressure must overcome: the hydrostatic pressure of the water at the depth of the tube's lower end, and the excess pressure required to form the spherical air bubble against surface tension.

1. Hydrostatic Pressure of the Water Column:
The lower end of the capillary tube is at a depth of $h=12\text{cm}$ below the water surface.
The hydrostatic pressure at this depth is given by:
$P_{\text{hydrostatic}}=h\rho g$
where $\rho$ is the density of water and $g$ is the acceleration due to gravity. This pressure is equivalent to a $12\text{cm}$ column of water.

2. Excess Pressure to Form the Bubble:
The excess pressure inside a spherical bubble of radius $r$ (equal to the radius of the capillary tube) in water is given by:
$P_{\text{excess}}=\frac{2T}{r}$
where $T$ is the surface tension of water.

We are given that water rises to a height of $h_0=3\text{cm}$ in the same capillary tube due to capillary action. The height of capillary rise is given by the formula:
$h_0=\frac{2T}{r\rho g}$
Rearranging this formula gives the excess pressure in terms of the water column height:
$P_{\text{excess}}=\frac{2T}{r}=h_0\rho g$
This excess pressure is equivalent to a $3\text{cm}$ column of water.

3. Total Required Mouth Pressure:
The total mouth pressure $P$ (gauge pressure) required to blow the bubble must overcome both the hydrostatic pressure and the capillary excess pressure:
$P=P_{\text{hydrostatic}}+P_{\text{excess}}$
$P=h\rho g+h_0\rho g=(h+h_0)\rho g$
Substituting the given values:
$P=(12+3)\text{cm of water column}=15\text{cm of water column}$