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 ........ $\text{cm}$.

$A$ capillary tube of radius '$r$' is immersed in water and water rises to a height of '$h$'. The mass of water in the capillary tube is $5 \times 10^{-3} \ kg$. The same capillary tube is now immersed in a liquid whose surface tension is $\sqrt{2}$ times the surface tension of water. The angle of contact between the capillary tube and this liquid is $45^{\circ}$. The mass of liquid which rises into the capillary tube now is (in $kg$):

In a capillary tube experiment, a vertical $30 \text{ cm}$ long capillary tube is dipped in water. Water rises up to a height of $10 \text{ cm}$ due to capillarity. If this experiment is conducted in a freely falling elevator, then the length of the water column becomes

The heat evolved for the rise of water when one end of the capillary tube of radius $r$ is immersed vertically into water is (Assume surface tension $= T$ and density of water to be $\rho$)

Two capillary tubes of length $L$ and $2L$ and of radius $R$ and $2R$ are connected in series. The net rate of flow of fluid through them will be (given rate of flow through a single capillary,$X = \frac{\pi P R^4}{8 \eta L}$)

$A$ long capillary tube of mass $\pi \,g$,radius $2\,mm$ and negligible thickness,is partially immersed in a liquid of surface tension $0.1\,N/m$. Take the angle of contact as zero and neglect the buoyant force of the liquid. Find the force required to hold the tube vertically. $(g = 10\,m/s^2)$