JEE Main & Advanced Physics NLM, Friction, Circular Motion JEE PYQ - NLM Friction Circular Motion

  • question_answer
    A particle of mass m is acted upon by a force F given by the empirical law \[F=\frac{R}{{{t}^{2}}}V(t)\].If this law is to be tested experimentally by observing the motion starting from rest, the best way is to plot:    [JEE ONLINE 10-04-2016]

    A) log v(t) against t

    B) v(t) against \[{{t}^{2}}\]

    C) log v(t) against\[\frac{1}{{{t}^{2}}}\]

    D) log v(t) against \[\frac{1}{t}\]

    Correct Answer: D

    Solution :

    [d] \[m\frac{dV}{dt}=\frac{R}{{{t}^{2}}}V\]
                \[\Rightarrow m\frac{dv}{v}=R\frac{dt}{{{t}^{2}}}\]          
                \[\Rightarrow \int\limits_{{{V}_{1}}}^{{{V}_{2}}}{\frac{dV}{V}}=R\int\limits_{{{t}_{1}}}^{{{t}_{2}}}{\frac{dt}{{{t}^{2}}}}\]
                \[\left. \Rightarrow \ell n\left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)=\frac{-R}{t} \right|_{{{t}_{1}}}^{{{t}_{2}}}\]
                \[\Rightarrow m\ell n\left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)=\frac{-R}{t}\left( \frac{1}{{{t}_{2}}}-\frac{1}{{{t}_{2}}} \right)\]
                \[\log V\,\,vs\,\,\frac{1}{t}\] will be a st. line curve

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