BCECE Engineering BCECE Engineering Solved Paper-2001

  • question_answer
    The work done to form a layer of soap solution of size \[10\times 10\,cm\] will be: (surface tension of soap solution is \[3\times {{10}^{-2}}\,N/m\])

    A)  \[\text{6 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-2}}}\text{J}\]                           

    B) \[\text{6 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-4}}}\text{J}\]            

    C)                        \[\text{3 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-2}}}\text{J}\]            

    D)         \[\text{3 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-4}}}\text{J}\]

    Correct Answer: B

    Solution :

    Key Idea: Surface tension T can be defined as the work done informing the soap film of unit surface area. Surface tension of a soap film is written as \[\text{T}\,\text{=}\,\frac{\text{work}\,\text{done}}{\text{area}}\]                 or            \[\text{work}\,\text{done}\,\text{=}\,\text{T}\,\text{ }\!\!\times\!\!\text{ }\,\,\text{area}\] Here,     \[=2\times 10\times 10\,c{{m}^{2}}\,=2\times {{10}^{-2}}\,{{m}^{2}},\] \[T=3\times {{10}^{-2}}N/m\]                 Hence, work done \[=3\times {{10}^{-2}}\,\times 2\times {{10}^{-2}}\] \[=6\times {{10}^{-4}}J\]


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