$A$ vertical glass capillary tube of radius $r$ open at both ends contains some water (surface tension $T$ and density $\rho$). If $L$ be the length of the water column,then:

The correct relation for the capillary rise $h$ in a tube of radius $r$ is:

One end of a capillary tube is dipped in water,the rise of water column is $h$. The upward force of $98 \text{ dyne}$ due to surface tension is balanced by the force due to the weight of the water column. The inner circumference of the capillary is (surface tension of water $= 7 \times 10^{-2} \text{ Nm}^{-1}$) (in $\text{ cm}$)

$A$ capillary tube of radius $r$ is initially just vertically completely immersed inside a liquid of angle of contact $0^{\circ}$. If the tube is slowly raised,then the relation between the radius of curvature of the meniscus inside the capillary tube and the displacement $(h)$ of the tube can be represented by:

Water rises to a height of $10$ cm in a capillary tube and mercury falls to a depth of $3.112$ cm in the same capillary tube. If the density of mercury is $13.6 \text{ g/cm}^3$ and the angle of contact for mercury is $135^o$,the ratio of surface tension of water and mercury is (Assume density of water = $1 \text{ g/cm}^3$ and angle of contact for water = $0^o$)

Water rises in a vertical capillary tube up to a height of $2.0\ cm$. If the tube is inclined at an angle of $60^o$ with the vertical,then up to what length will the water rise in the tube? (in $cm$)