JEE Main & Advanced Physics Fluid Mechanics, Surface Tension & Viscosity / द्रव यांत्रिकी, भूतल तनाव और चिपचिपापन Question Bank Surface Energy

  • question_answer
    Work done in splitting a drop of water of 1 mm radius into 106 droplets is (Surface tension of water \[=72\times {{10}^{-3}}J/{{m}^{2}})\] [MP PET/PMT 1988; CPMT 1989; RPET 2001]

    A)                     \[9.58\times {{10}^{-5}}\,J\]           

    B)             \[8.95\times {{10}^{-5}}\,J\]

    C)             \[5.89\times {{10}^{-5}}\,J\]           

    D)              \[5.98\times {{10}^{-6}}J\]

    Correct Answer: B

    Solution :

                    Work done in splitting a water drop of radius R into n drops of equal size\[=4\pi {{R}^{2}}T({{n}^{1/3}}-1)\]             \[=4\pi \times {{({{10}^{-3}})}^{2}}\times 72\times {{10}^{-3}}\times ({{10}^{6/3}}-1)\]             \[=4\pi \times {{10}^{-6}}\times 72\times {{10}^{-3}}\times 99=8.95\times {{10}^{-5}}J\]


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