Water rises to a height of $20 \ mm$ in a capillary tube. If the radius of the capillary tube is made $\frac{1}{3}rd$ of its previous value,what is the new height of the capillary rise (in $mm$)?

Two capillary tubes $P$ and $Q$ are dipped vertically in water. The height of water level in capillary tube $P$ is $\frac{2}{3}$ of the height in capillary tube $Q$. The ratio of their diameters is

Two parallel glass plates are dipped partly in a liquid of density $d$ while keeping them vertical. If the distance between the plates is $x$, the surface tension of the liquid is $T$, and the angle of contact is $\theta$, then the rise of the liquid between the plates due to capillarity will be:

$A$ capillary tube made of glass with a radius of $0.15\, mm$ is dipped vertically in a beaker filled with methylene iodide (surface tension $= 0.05\, N m^{-1}$,density $= 667\, kg m^{-3}$),which rises to a height $h$ in the tube. It is observed that the two tangents drawn from the liquid-glass interfaces (from opposite sides of the capillary) make an angle of $60^{\circ}$ with one another. Then $h$ is close to $...... m$ $(g = 10\, m s^{-2})$

Two capillaries made of the same material have radii $r_1 = 1 \, mm$ and $r_2 = 2 \, mm$. If the rise of the liquid in the first capillary $(r_1 = 1 \, mm)$ is $30 \, cm$,then the rise in the other capillary will be ........ $cm$.

When a long glass capillary tube of radius $0.015 \; cm$ is dipped in a liquid,the liquid rises to a height of $15 \; cm$ within it. If the contact angle between the liquid and glass is close to $0^{\circ}$,the surface tension of the liquid,in $milliNewton \; m^{-1}$ is $.....$
$[\rho_{\text{liquid}} = 900 \; kg \; m^{-3}, g = 10 \; ms^{-2}]$ (Give answer in closest integer)