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JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Afternoon)

  • question_answer
    A solid sphere, of radius R acquires a terminal velocity\[{{\text{v}}_{1}}\]when falling (due to gravity) through a viscous fluid having a coefficient of viscosity \[\eta \]. The sphere is broken into 27 identical solid spheres. If each of these spheres acquires a terminal velocity, \[{{\text{v}}_{2}},\]when falling through the same fluid, the ratio \[\text{(}{{\text{v}}_{1}}/{{\text{v}}_{2}})\]equals : [JEE Main 12-4-2019 Afternoon]

    A) 1/27                

    B) 1/9

    C) 27                               

    D) 9

    Correct Answer: D

    Solution :

    We have \[{{V}_{T}}=\frac{2}{9}\frac{{{r}^{2}}}{\eta }({{\rho }_{0}}-{{\rho }_{\ell }})g\Rightarrow {{V}_{T}}\propto {{r}^{2}}\] since mass of the sphere will be same \[\therefore \]\[\rho \frac{4}{3}\pi {{R}^{3}}=24.\frac{4}{3}\pi {{r}^{3}}\rho \Rightarrow r=\frac{R}{3}\] \[\therefore \]\[\frac{{{\text{v}}_{\text{1}}}}{{{\text{v}}_{\text{2}}}}=\frac{{{R}^{2}}}{{{r}^{2}}}=9\]


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