A glass capillary of radius 0.35 mm is inclined at $60^{\circ}$ with the vertical in water. The length of water column in the capillary tube is (surface tension of water $=7 \times 10^{-2} \mathrm{Nm}^{-1}$, acceleration due to gravity $=10 \mathrm{~m} / \mathrm{s}^2, \cos 0^{\circ}=1$, $\cos 60^{\circ}=0.5$, density of water $=1 \mathrm{gram} / \mathrm{cm}^3$ )

At two points on a horizontal tube of varying cross-section the radii are 1 cm and $\mathbf{0 . 4 ~ c m}$, velocities of fluid are $\mathrm{V}_1, \mathrm{~V}_2$ and pressure difference ( $P_1-P_2$ ) between these points is 4.9 cm of water. The value of $\sqrt{V_2^2-V_1^2}$ is ( $\mathrm{g}=$ acceleration due to gravity $\mathrm{g}=980 \mathrm{~cm} / \mathrm{s}^2$ )

Water rises up to height ' $x$ ' in a capillary tube immersed vertically in water. When the whole arrangement is taken to a depth 'd' in a mine, the water level rises height ' $Y$ '. If ' $R$ ' is the radius of earth then the ratio $(\mathrm{Y} / \mathrm{x})$ is