question_answer 19)

What is the pressure inside a drop of mercury of radius 3.0 mm at room temperature? Surface tension of mercury at that temperature (20C) is \[{{10}^{8}}\times \pi {{r}^{2}}={{10}^{8}}\times \left( 22/7 \right)\times {{\left( 1\cdot 5\times {{10}^{-2}} \right)}^{2}}=7\cdot 07\times {{10}^{4}}N\]. The atmospheric pressure is \[1.01\times {{10}^{5}}\,Pa\] . Also give the excess pressure inside the drop.

Answer:

Here, \[r=3.0\,mm\,\,=3\times {{10}^{-3}}\,m;\] \[190\times {{10}^{9}}\] Excess of pressure inside the drop of mercury is given by, \[N{{m}^{-2}}\] Total pressure inside the drop \[\text{Y=}\frac{\text{4F/ }\!\!\pi\!\!\text{ }{{\text{D}}^{\text{2}}}}{\text{strain}}\]