KVPY Sample Paper KVPY Stream-SX Model Paper-4

  • question_answer
    A gas is a mixture of two parts by volume of hydrogen and one part by volume of nitrogen. If the velocity of sound in hydrogen at \[0{}^\circ C\]is \[1300m/s\] then the velocity of sound in the gaseous mixture at \[27{}^\circ \operatorname{C}\]will be

    A) \[951m/sec\]                  

    B) \[159m/sec\]

    C) \[591m/sec\]      

    D) \[519m/sec\]

    Correct Answer: C

    Solution :

    we have \[\frac{{{v}_{m}}}{{}^{v}H}=\sqrt{\frac{{{\rho }_{H}}}{{{\rho }_{m}}}}\] ? (1)
    Here, \[{{\rho }_{m}}=\frac{{{m}_{H}}+{{m}_{N}}}{{{V}_{H}}+{{V}_{N}}}\]
    \[=\frac{{{\rho }_{H}}{{V}_{H}}+{{\rho }_{N}}{{V}_{N}}}{{{V}_{H}}+{{V}_{N}}}\]
    \[=\frac{{{\rho }_{H}}{{V}_{H}}\left( 1+\frac{{{\rho }_{N}}}{{{\rho }_{H}}}\times \frac{{{V}_{N}}}{{{V}_{H}}} \right)}{{{V}_{H}}\left( 1+\frac{{{V}_{N}}}{+{{V}_{H}}} \right)}\]
    \[\therefore \frac{{{\rho }_{m}}}{{{\rho }_{H}}}=\frac{1+\frac{28}{2}\times \frac{1}{2}}{1+\frac{1}{2}}=\frac{16}{3}\]
    Now from equation (i), we have \[{{[{{v}_{m}}]}_{0}}={{v}_{H}}=\sqrt{\frac{3}{16}}\]
    \[{{[{{v}_{m}}]}_{27}}={{({{v}_{m}})}_{0}}\left( 1+\frac{t}{546} \right)=591m/s\]

You need to login to perform this action.
You will be redirected in 3 sec spinner