JEE Main & Advanced Physics Fluid Mechanics, Surface Tension & Viscosity / द्रव यांत्रिकी, भूतल तनाव और चिपचिपापन Question Bank Self Evaluation Test - Mechanical Properties of Fluids

  • question_answer
    A water film is formed between two straight parallel wires of 10 cm length 0.5 cm apart. If the distance between wires is increased by 1 mm. What will be the work done? (surface tension of water \[=72dyne/cm\] )

    A) 36 erg

    B) 288 erg

    C) 144 erg

    D) 72 erg

    Correct Answer: C

    Solution :

    [c] Work done = Surface tension \[\times \] increase in area of the film \[W=S\times \Delta A\] \[Increase\text{ }in\text{ }area\text{ }=Final\text{ }area\text{ }-\text{ }initial\text{ }area\]\[=10\times (0.5+0.1)-10\times 0.5=1c{{m}^{2}}\] \[\therefore \,\,\,W=72\times 2\times 1=144erg\] [\[\therefore \] There are 2 free surfaces; \[\therefore \,\,\Delta A=2\times 1\]].


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