$A$ liquid rises to a height of $2.4 \ cm$ in a glass capillary $P$. Another glass capillary $Q$ having a diameter $80\%$ of capillary $P$ is immersed in the same liquid. The rise of liquid in capillary $Q$ is (in $cm$)

$A$ capillary tube of radius $0.1 \ mm$ is partly dipped in water (surface tension $70 \ dyn/cm$ and glass-water contact angle $\simeq 0^{\circ}$) inclined at $30^{\circ}$ with the vertical. The length of water risen in the capillary is . . . . . . $cm$. (Take $g = 980 \ cm/s^2$)

Two capillary tubes of the same diameter are kept vertically in two different liquids whose densities are in the ratio $4:3$. The rise of liquid in two capillaries is $h_1$ and $h_2$ respectively. If the surface tensions of liquids are in the ratio $6:5$,the ratio of heights $\left(\frac{h_1}{h_2}\right)$ is (Assume that their angles of contact are same).

Water rises to a height of $16.3 \,cm$ in a capillary tube. If the tube is cut at a height of $12 \,cm$ above the water level,what will happen?

Water rises up to a height of $4 \,cm$ in a capillary tube. The lower end of the capillary tube is at a depth of $8 \,cm$ below the water level. The mouth pressure required to blow an air bubble at the lower end of the capillary will be '$X$' $cm$ of water,where $X$ is equal to

Liquid $A$ rises to a height of $10 \ cm$ in a capillary tube and liquid $B$ falls to a depth of $2 \ cm$ in the same tube. The densities of $A$ and $B$ are $1 \ g/cm^3$ and $10 \ g/cm^3$ respectively. The contact angles of $A$ and $B$ with the tube are $0^{\circ}$ and $135^{\circ}$ respectively. If the surface tensions of $A$ and $B$ are $S_A$ and $S_B$,then the ratio $\frac{S_B}{S_A}$ is: