Question

The surface tension of a soap solution is $ 2\times {{10}^{-2}} $ N/m. To blow a bubble of radius 2 cm from 1 cm, the work done is

• A. $ 1.5\pi \times {{10}^{-4}}J $
• B. $ 1.2\pi \times {{10}^{-4}}J $
• C. $ 0.3\pi \times {{10}^{-4}}J $
• D. $ 0.6\pi \times {{10}^{-4}}J $

Answer 1

Answer: (B)

$ W=2T\Delta A $
$ =2T\times 4\pi (r_{1}^{2}-r_{2}^{2}) $
$ =2\times 0.05\times 4\pi \times (4\times {{10}^{4}}-1\times {{10}^{4}}) $
$ W=1.2\pi \times {{10}^{-4}}J $