JEE Main & Advanced JEE Main Paper Phase-I (Held on 07-1-2020 Evening)

  • question_answer
    Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently the mean collision time between the gas molecule changes from \[{{\tau }_{1}}\] to\[{{\tau }_{2}}\]. If \[\frac{{{C}_{P}}}{{{C}_{v}}}=\gamma \]for this gas then a good estimate for \[\frac{{{\tau }_{2}}}{{{\tau }_{1}}}\]is given by                                                               [JEE MAIN Held on 07-01-2020 Evening]

    A) \[{{\left( \frac{1}{2} \right)}^{\frac{\gamma +1}{2}}}\]  

    B) \[{{\left( \frac{1}{2} \right)}^{\gamma }}\]

    C) \[2\]                 

    D) \[\frac{1}{2}\]

    Correct Answer: A , B , C , D

    Solution :

    (BONUS) \[\tau \propto \frac{1}{n<v>},\,<v>\,\propto \sqrt{T}\]             \[\Rightarrow \,\,\,\tau \propto \frac{1}{n\sqrt{T}}\Rightarrow \,\frac{{{\tau }_{2}}}{{{\tau }_{1}}}=\frac{{{n}_{1}}}{{{n}_{2}}}\sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}}\]             \[=2\sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}}\]             \[{{T}_{1}}{{V}_{1}}^{\gamma -1}={{T}_{2}}{{(2{{V}_{1}})}^{\gamma -1}}\Rightarrow \,\frac{{{T}_{1}}}{{{T}_{2}}}={{2}^{\gamma -1}}\]             \[\Rightarrow \,\,\frac{{{\tau }_{2}}}{{{\tau }_{1}}}\,=2\times {{2}^{\frac{(\gamma -1)}{2}}}=2{{\,}^{\left( \frac{\gamma +1}{2} \right)}}\]   NOTE: Answer does not match with given options.

    Solution :

    (BONUS) \[\tau \propto \frac{1}{n<v>},\,<v>\,\propto \sqrt{T}\]             \[\Rightarrow \,\,\,\tau \propto \frac{1}{n\sqrt{T}}\Rightarrow \,\frac{{{\tau }_{2}}}{{{\tau }_{1}}}=\frac{{{n}_{1}}}{{{n}_{2}}}\sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}}\]             \[=2\sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}}\]             \[{{T}_{1}}{{V}_{1}}^{\gamma -1}={{T}_{2}}{{(2{{V}_{1}})}^{\gamma -1}}\Rightarrow \,\frac{{{T}_{1}}}{{{T}_{2}}}={{2}^{\gamma -1}}\]             \[\Rightarrow \,\,\frac{{{\tau }_{2}}}{{{\tau }_{1}}}\,=2\times {{2}^{\frac{(\gamma -1)}{2}}}=2{{\,}^{\left( \frac{\gamma +1}{2} \right)}}\]   NOTE: Answer does not match with given options.

    Solution :

    (BONUS) \[\tau \propto \frac{1}{n<v>},\,<v>\,\propto \sqrt{T}\]             \[\Rightarrow \,\,\,\tau \propto \frac{1}{n\sqrt{T}}\Rightarrow \,\frac{{{\tau }_{2}}}{{{\tau }_{1}}}=\frac{{{n}_{1}}}{{{n}_{2}}}\sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}}\]             \[=2\sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}}\]             \[{{T}_{1}}{{V}_{1}}^{\gamma -1}={{T}_{2}}{{(2{{V}_{1}})}^{\gamma -1}}\Rightarrow \,\frac{{{T}_{1}}}{{{T}_{2}}}={{2}^{\gamma -1}}\]             \[\Rightarrow \,\,\frac{{{\tau }_{2}}}{{{\tau }_{1}}}\,=2\times {{2}^{\frac{(\gamma -1)}{2}}}=2{{\,}^{\left( \frac{\gamma +1}{2} \right)}}\]   NOTE: Answer does not match with given options.

    Solution :

    (BONUS) \[\tau \propto \frac{1}{n<v>},\,<v>\,\propto \sqrt{T}\]             \[\Rightarrow \,\,\,\tau \propto \frac{1}{n\sqrt{T}}\Rightarrow \,\frac{{{\tau }_{2}}}{{{\tau }_{1}}}=\frac{{{n}_{1}}}{{{n}_{2}}}\sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}}\]             \[=2\sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}}\]             \[{{T}_{1}}{{V}_{1}}^{\gamma -1}={{T}_{2}}{{(2{{V}_{1}})}^{\gamma -1}}\Rightarrow \,\frac{{{T}_{1}}}{{{T}_{2}}}={{2}^{\gamma -1}}\]             \[\Rightarrow \,\,\frac{{{\tau }_{2}}}{{{\tau }_{1}}}\,=2\times {{2}^{\frac{(\gamma -1)}{2}}}=2{{\,}^{\left( \frac{\gamma +1}{2} \right)}}\]   NOTE: Answer does not match with given options.


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