• # question_answer The work done against surface tension in blowing a soap bubble from a radius 10 to 20 cm, if the surface tension of solution is $25\times {{10}^{-3}}N/m,$ is A) $3\pi \times {{10}^{-3}}J$             B) $6\pi \times {{10}^{-3}}J$C) $2\pi \times {{10}^{-3}}J$             D) $4\pi \times {{10}^{-3}}J$

Initial total surface area${{A}_{1}}=2\times 4\pi {{(v)}^{2}}{{m}^{2}}.$ Final total surface area ${{A}_{2}}=2\times 4\pi {{(0.2)}^{2}}{{m}^{2}}.$ Work done against surface tension,$W=S\,({{A}_{2}}-{{A}_{1}})$ $\therefore$$W=25\times {{10}^{-3}}\times 8\pi (0.04-0.01)=6\pi \times {{10}^{-3}}J$ Hence, the correction option is [b].