JEE Main & Advanced JEE Main Paper (Held On 10 April 2016)

  • 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 Main Online Paper (Held On 10 April 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 :

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


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