$A$ $20 cm$ long capillary tube is dipped vertically in water and the liquid rises up to $10 cm$. If the entire system is kept in a freely falling platform, the length of the water column in the tube will be (in $cm$)

Two capillary tubes $A$ and $B$ of the same internal diameter are kept vertically in two different liquids whose densities are in the ratio $4:3$. If the surface tensions of these two liquids are in the ratio $6:5$,then the ratio of rise of liquid in capillary $A$ to that in $B$ is (assume their angles of contact are nearly equal).

Radius of a capillary is $2 \times 10^{-3} \,m$. $A$ liquid of weight $6.2 \times 10^{-4} \,N$ may remain in the capillary. Then the surface tension of the liquid will be $:-$

$A$ capillary tube of radius $r$ is immersed in water and water rises in it to a height $h$. The mass of the water in the capillary tube is $m$. Another capillary of radius $2r$ is immersed in water. The mass of water that will rise in this tube is

When a capillary tube is kept in water, water $......$ in the capillary, while when it is kept in mercury, the mercury $......$ in the capillary. (Choose the correct words: $rising up$ / $depressed$)

$A$ glass capillary tube is in the shape of a truncated cone with an apex angle $\alpha$ so that its two ends have cross sections of different radii. When dipped in water vertically,water rises in it to a height $h$,where the radius of its cross section is $b$. If the surface tension of water is $S$,its density is $\rho$,and its contact angle with glass is $\theta$,the value of $h$ will be ($g$ is the acceleration due to gravity).