A glass tube of circular cross section is closed at one end. This end is weighted and the tube floats vertically in water, heavy end down. The length (in $cm$ ) of the tube below the water surface is ___ . (Given outer radius of the tube is $0.14 cm$ , mass of weighted tube is $0 .2 g$ , surface tension of water $73 dyne/cm$ and $g=980 cm{s}^{-2}$ .)

Let $l$ be the length of the tube inside water. The forces acting on the tube are : Upthrust of water acting upward $={\mathrm{\pi r}}^{2}l \times l \times 980$

$=\frac{22}{7}\times {\left(0.14\right)}^{2}l\times 980=60.38l dyne$

Weight of the system acting downwards $=mg$$= 0.2 \times 980 = 196 dyne$

Force of surface tension acting downward $= 2\mathrm{\pi }rT = 64.24 dyne$

Since the tube is in equilibrium, the upward force is balanced by the downward forces. That is, $60.3 68l= 196 + 64.24 = 260.24 I = 4.31 cm$