Question

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, then the mass of water that rises in the capillary tube will be:

• A. $M$
• B. $2M$
• C. $\frac{M}{2}$
• D. $4M$

Answer 1

The correct option is B $2M$

Mass of liquid in capillary tube is,
$M=\pi R^{2}H\times \rho$-----------(1)
The rise of liquid in capillary tube,
$H=\frac{2T\cos \theta }{\rho Rg}$------------(2)

Putting value of H from equation 2 to equation 1

We get,
$\Rightarrow M=\frac{2\pi R^2\rho T\cos \theta }{\rho Rg}$
For the same liquid and capillary tube $T,\theta ,\rho$ are constant.
$\Rightarrow M\propto R.$
Hence, if radius is doubled, then mass will become two times.
$\therefore M_{\text{new}}=2M$