A capillary tube of radius R is immersed in water and water rises in it to a height H. Mass of water in the capillary tube is M. If the radius of the tube is doubled, mass of water that will rise in the capillary tube will now be

The correct option is 2M.

Let's understand the physical principles governing the rise of liquid in a capillary tube step-by-step.

When a capillary tube of radius $R$ is immersed in water, the water rises to a height $H$ due to surface tension. According to Jurin's Law, the height $H$ to which a liquid rises in a capillary tube is given by the formula:
$H=\frac{2T\cos \theta }{R\rho g}$
where:
- $T$ is the surface tension of water,
- $\theta$ is the angle of contact,
- $\rho$ is the density of water, and
- $g$ is the acceleration due to gravity.

From this relation, we can see that for a given liquid and tube material, the height of rise is inversely proportional to the radius of the tube:
$H\propto \frac{1}{R}$
This means:
$H\times R=\text{constant}$

Now, let's look at the mass $M$ of water in the capillary tube. Assuming the tube has a circular cross-section, the volume of the water column of height $H$ is:
$V=\pi R^{2}H$

The mass $M$ is the product of volume $V$ and the density $\rho$:
$M=\rho V=\rho \pi R^{2}H$

Since $\rho$ and $\pi$ are constants, the mass is proportional to the product $R^{2}H$:
$M\propto R^{2}H$

We can rewrite this expression as:
$M\propto R\times (R\times H)$

Since we established from Jurin's Law that $R\times H$ is a constant, the mass $M$ of water in the capillary tube is directly proportional to the radius $R$ of the tube:
$M\propto R$

Let $M_1=M$ be the initial mass when the radius is $R_1=R$.
When the radius is doubled, the new radius becomes $R_2=2R$.

Let the new mass be $M_2$. Using the proportionality relation:
$\frac{M_2}{M_1}=\frac{R_2}{R_1}$
Substituting the values:
$\frac{M_2}{M}=\frac{2R}{R}=2$
$M_2=2M$

Therefore, when the radius of the tube is doubled, the mass of water rising in the capillary tube is also doubled, which is $2M$.