Water rises in a capillary tube to a certain height such that the upward force due to surface tension is balanced by $75 \times 10^{-4} \, N$ force due to the weight of the liquid. If the surface tension of water is $6 \times 10^{-2} \, N m^{-1}$,the inner circumference of the capillary must be

Two capillary tubes of radii $R$ and $2R$ are dipped in water. The water rises to heights $h_1$ and $h_2$ respectively in the two tubes. If $m_1$ and $m_2$ be the masses of water ascended in the tubes,then $m_1: m_2=$

Water rises in a capillary tube to a height such that the upward force of surface tension balances the downward force of $75 \times 10^{-4} \ N$ due to the weight of the water. If the surface tension of water is $12 \times 10^{-2} \ N/m$,the internal circumference of the capillary must be:

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: $rising up$ / $depressed$)

$A$ 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 $kg$):

Water rises up to a height $h$ in a capillary tube immersed vertically in water. When this whole arrangement is taken to a depth $d$ in a mine,the water level rises up to a height $h^{\prime}$. If $R$ is the radius of the earth,then the ratio $\frac{h}{h^{\prime}}$ is