Water rises to a height of $20 \,mm$ in a capillary tube of cross-sectional area $A$. If the area of cross-section of the tube is made $\frac{A}{4}$, then water will rise to a height of (in $\,cm$)

$A$ 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 $kg$):

$A$ capillary tube is immersed vertically in water,and the water rises to a height $h_1$ on the Earth's surface. When this arrangement is taken into a mine of depth $d$ below the Earth's surface,the height of the water column is $h_2$. If $R$ is the radius of the Earth,the ratio $\frac{h_2}{h_1}$ is:

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 $\rho$)

$A$ 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. $(g = 10\,m/s^2)$

When one end of a capillary tube is dipped in water,the height of water column is $h$. The upward force of $105 \text{ dyne}$ due to surface tension is balanced by the force due to the weight of water column. The inner circumference of the capillary tube is (Surface tension of water $= 7 \times 10^{-2} \text{ N/m}$) (in $\text{ cm}$)