question_answer 49)

                  Two mercury droplets of radii 0.1 cm and 0.2 cm collapse into one single drop. What amount of energy is released? The surface tension of mercury \[T=435.5\times {{10}^{-3}}\,N\,{{m}^{-1}}\].                

Answer:

                  Volume of 2 droplets = volume of a single drop                 i.e. \[\frac{4}{3}\,\pi \,r_{1}^{3}\,\,+\,\frac{4}{3}\,\pi \,r_{2}^{3}\,=\,\frac{4}{3}\,\pi {{R}^{3}}\]                 or \[{{R}^{3}}=(r_{1}^{3}+r_{2}^{2})={{10}^{-3}}+8\times {{10}^{-3}}\]                 \[\therefore \] \[R=\,2.08\,\times \,{{10}^{-1}}\,=0.208\,=\,0.21\,\,cm\]                 Decrease in surface area                 \[=\ 4\times \,3.14\,\,[0.0441-0.05]=-0.074\,\,c{{m}^{2}}\,\]                 =\[=-0.074\times {{10}^{-4}}\,{{m}^{2}}\]                 \[\therefore \] Energy released \[=T\times \] decrease in surface area                 \[=-435.5\times {{10}^{-3}}\times 0.074\times {{10}^{-4}}\]                         \[=-32.23\times {{10}^{-7}}\,J\]