The angle of contact between glass and water is $0^\circ$ and it rises in a capillary up to $6 \text{ cm}$ when its surface tension is $70 \text{ dynes/cm}$. Another liquid of surface tension $140 \text{ dynes/cm}$,angle of contact $60^\circ$ and relative density $2$ will rise in the same capillary by ........ $\text{cm}$.

According to Poiseuille's law,the pressure drop per unit length required to overcome viscous forces is $\Delta P = \frac{8 \eta v}{r^2}$,where $r$ is the radius of the cross-section,$v$ is the fluid velocity,and $\eta$ is the coefficient of viscosity. $A$ capillary tube of radius $a$ is dipped in a liquid of density $\rho$,surface tension $T$,and coefficient of viscosity $\eta$. The liquid starts rising in it so that its height $h(t)$ is a function of time $t$. The resulting rate of change of the momentum of the liquid column in the capillary (taking vertically up to be the positive direction and the contact angle to be close to $0^{\circ}$) is $-\pi a^2 \rho gh + F$. Then $F$ is ($g$ is the acceleration due to gravity):

During the capillary rise of a liquid in a capillary tube,the surface of contact that remains constant is of

Three liquids have the same surface tension and densities $\rho_1, \rho_2$,and $\rho_3$ $(\rho_1 > \rho_2 > \rho_3)$. In three identical capillaries,the rise of the liquid is the same. The corresponding angles of contact $\theta_1, \theta_2$,and $\theta_3$ are related as:

Two capillary tubes of different diameters are dipped in water. The rise of water is

Two capillary tubes of radii $0.2 \ cm$ and $0.4 \ cm$ are dipped in the same liquid. The ratio of the heights to which the liquid will rise in the tubes is: