When a capillary tube is immersed in water vertically,water rises to a height $h$ inside the tube. If the radius of another capillary tube is $\frac{1}{3}$ that of the previous,the height to which water will rise in this tube is:

$A$ $U$-tube with limbs of diameters $5\, mm$ and $2\, mm$ contains water of surface tension $7 \times 10^{-2} \, N/m$. The angle of contact is zero and the density of water is $10^3 \, kg/m^3$. If $g = 10 \, m/s^2$,then the difference in the liquid levels of the two limbs is:

When a capillary tube of radius '$r$' is immersed in water, the rise of water is up to height '$h$'. The mass of water in the capillary tube is '$m$'. When another capillary tube of radius '$xr$' is immersed in water, the mass of water that will rise in this tube is:

$A$ $U$-tube is such that the diameter of one limb is $0.4\,mm$ and that of the other is $d\,mm$. If the surface tension of water contained in the tube is $0.07\,N/m$ and the difference in the levels of liquid in the limbs is $3.6\,cm$,then the value of $d$ is:

By inserting a capillary tube up to a depth $l$ in water,the water rises to a height $h$. If the lower end of the capillary is closed inside water,the capillary is taken out,and then the closed end is opened,to what height will the water remain in the tube?

$A$ vertical capillary tube is filled with water up to the top after closing its lower end with a finger. If the finger is removed,we shall observe that: ($T = 70 \,\, dyne/cm$,radius of capillary $r = 1 \,\, mm$ and $g = 980 \,\, cm/sec^2$)