When one end of the capillary is dipped in water, the height of water column is ' $h$ '. The upward force of 108 dyne due to surface tension is balanced by the force due to the weight of water column. The inner circumference of the capillary is (surface tension of water $=7.2 \times 10^{-2} \mathrm{~N} / \mathrm{m}$ )

A capillary tube when immersed vertically in water, the rise of water column is upto height $h_1$ on earth's surface. When this arrangement is taken into a mine of depth 'd', below earth's surface, the height of the water column is $\mathrm{h}_2$. If R is the radius of the earth, the ratio $\frac{\mathrm{h}_2}{\mathrm{~h}_1}$ is

125 small water drops of same size fall through air with constant velocity $4 \mathrm{~cm} / \mathrm{s}$. They coalesce to form a big drop. The terminal velocity of the big drop is

Let $R_1, R_2$ and $R_3$ be the radii of three mercury drops. A big mercury drop is formed from them under isothermal conditions. The radius of the resultant drop is