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).

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: $rising up$ / $depressed$)

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 $\text{ cm}$)

Water rises up to a height $h$ in a capillary tube of a certain diameter. This capillary tube is replaced by a similar tube of half the diameter. Now,the water will rise to a height of:

Two narrow bores of diameters $3.0 \;mm$ and $6.0 \;mm$ are joined together to form a $U$-tube open at both ends. If the $U$-tube contains water,what is the difference (in $mm$) in its levels in the two limbs of the tube? Surface tension of water at the temperature of the experiment is $7.3 \times 10^{-2} \;N m^{-1}$. Take the angle of contact to be zero and density of water to be $1.0 \times 10^{3} \;kg m^{-3}$ $(g = 9.8 \;m s^{-2})$.

Water rises in a vertical capillary tube up to a height of $2.0 \, cm$. If the tube is inclined at an angle of $60^{\circ}$ with the vertical,then up to what length will the water rise in the tube?