1 The surface tension of a liquid enables a metal needle to be at rest on the surface of the liquid. The greatest mass \(m\) of a needle of length \(a\) which can be supported in this way by a liquid of surface tension \(S\) is given by $$m = \frac { 2 S a } { g }$$ where \(g\) is the acceleration due to gravity.

1. Show that the dimensions of surface tension are \(\mathrm { MT } ^ { - 2 }\). The surface tension of water is 0.073 when expressed in SI units (based on kilograms, metres and seconds).
2. Find the surface tension of water when expressed in a system of units based on grams, centimetres and minutes. Liquid will rise up a capillary tube to a height \(h\) given by \(h = \frac { 2 S } { \rho g r }\), where \(\rho\) is the density of the liquid and
   \(r\) is the radius of the capillary tube. \(r\) is the radius of the capillary tube.
3. Show that the equation \(h = \frac { 2 S } { \rho g r }\) is dimensionally consistent.
4. Find the radius of a capillary tube in which water will rise to a height of 25 cm . (The density of water is 1000 in SI units.) When liquid is poured onto a horizontal surface, it forms puddles of depth \(d\). You are given that \(d = k S ^ { \alpha } \rho ^ { \beta } g ^ { \gamma }\) where \(k\) is a dimensionless constant.
5. Use dimensional analysis to find \(\alpha , \beta\) and \(\gamma\). Water forms puddles of depth 0.44 cm . Mercury has surface tension 0.487 and density 13500 in SI units.
6. Find the depth of puddles formed by mercury on a horizontal surface.