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    				| 1. | Find the excess pressure inside a mercury drop of radius 2.0 mm. The surface tension oif mercury `0.464Nm^-1`. | 
| Answer» The excess pressure inside the drop is P_2-P_1=2S/R` `=(2xx0.464Nm^-1)/(2.0xx10^-3m)=464Nm^-2` | |