KVPY Sample Paper KVPY Stream-SX Model Paper-22

  • question_answer
    A tiny spherical oil drop carrying a net charge \[q\] is balanced in still air with a vertical uniform electric field of strength\[\,\frac{81\pi }{7}\times {{10}^{5}}V{{m}^{-1}}\]. When the field is switched off, the drop is observed to fall with terminal velocity \[2\times {{10}^{-3}}m{{s}^{-1}}\] Given \[g=9.8\text{ }{{\operatorname{ms}}^{-2}}\], viscosity of the air \[=1.8\times \text{ }\]\[{{10}^{-5}}Ns\,{{m}^{-2}}\] and the density of oil \[=900\text{ }kg\,{{\operatorname{m}}^{-3}}\] the magnitude of \[q\] is

    A) \[1.6\times {{10}^{-19}}C\]

    B) \[2.2\times {{10}^{-19}}C\]

    C) \[4.8\times {{10}^{-19}}C\]

    D) \[8.0\times {{10}^{-19}}C\]

    Correct Answer: D

    Solution :

    When the electric field is on
    Force due to electric field= weight
    \[qE=\frac{4}{3}\pi {{R}^{3}}\rho g\]
    \[qE=\frac{4\pi {{R}^{3}}\rho g}{3E}...(i)\]
    When the electric field is switched off
    Weight=viscous drag force
    \[\operatorname{mg}=6\pi \eta R{{v}_{t}}\]
    \[\frac{4}{3}\pi {{R}^{3}}\rho g=6\pi \eta R{{v}_{t}}\]\[\therefore \,\,\,\,R=\sqrt{\frac{9\nu \eta {{v}_{t}}}{2\rho g}}...\left( \operatorname{ii} \right)\]
    From (i) and (ii), \[q=\frac{4}{3}\pi {{\left[ \frac{9\nu \eta {{v}_{t}}}{2\rho g} \right]}^{\frac{3}{2}}}\times \frac{\rho g}{E}\]\[=\frac{4}{3}\times \pi {{\left[ \frac{9\times 1.8\times {{10}^{-5}}\times 2\times {{10}^{-3}}}{2\times 900\times 9.8} \right]}^{\frac{3}{2}}}\]\[\times \frac{900\times 9.8\times 7}{81\pi \times {{10}^{5}}}\]

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