A) \[3\pi \times {{10}^{-3}}J\]
B) \[6\pi \times {{10}^{-3}}J\]
C) \[2\pi \times {{10}^{-3}}J\]
D) \[4\pi \times {{10}^{-3}}J\]
Correct Answer: B
Solution :
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].
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