• # question_answer If the work done in blowing a bubble of volume V is W, then the work done in Mowing a soap bubble of volume 2V will be A)  W                                B)  2W C)  $\sqrt{2}W$              D)  ${{4}^{1/3}}W$

From the definition of surface tension (T), the surface tension of a liquid is equal to the work (W) required to increase the surface area  of the liquid film by unity at constant temperature, $\therefore$    $W=T\times \Delta A$ Since, surface area of a sphere is $4\pi {{R}^{2}}$ and there are two free surfaces, we have $W=T\times \Delta A$        ?(i) and volume of sphere $=\frac{4}{3}\,\pi {{R}^{3}}$ i.e.,                  $V=\frac{4}{3}\,\pi {{R}^{3}}$ $\Rightarrow$                        $R={{\left( \frac{3V}{4\pi } \right)}^{1/3}}$         ?(ii) From Eqs. (i) and (ii), we get $W=T\times 8\pi \times {{\left( \frac{3V}{4\pi } \right)}^{2/3}}$ $\Rightarrow$            $W\propto \,{{V}^{2/3}}$ $\therefore$    ${{W}_{1}}\propto \,{{V}_{1}}^{2/3}$ and      ${{W}_{2}}\propto \,{{V}_{2}}^{2/3}$ $\therefore$    $\frac{{{W}_{2}}}{{{W}_{1}}}=\,{{\left( \frac{2V}{V} \right)}^{2/3}}$ $\Rightarrow$            ${{W}_{2}}={{2}^{2/3}}{{W}_{1}}={{4}^{1/3}}W$