J & K CET Engineering J and K - CET Engineering Solved Paper-2014

  • question_answer An air bubble of \[2\text{ }c{{m}^{3}}\]rises from the bottom of a lake of \[32\text{ }m\]at a temperature of \[{{9}^{o}}C\]. When the bubble reaches the surface of the lake from the bottom of the lake, what volume does it grows for which temperature is \[{{30}^{o}}C\](assume \[g=10m/{{s}^{2}}\]and density\[\rho ={{10}^{3}}\text{ }kg/{{m}^{3}}\])

    A)  \[5.937\text{ }c{{m}^{3}}\]       

    B)  \[8.937\text{ }c{{m}^{3}}\]

    C)  \[~12.937\,\,c{{m}^{3}}\]       

    D)  \[16.937\,\,c{{m}^{3}}\]

    Correct Answer: B

    Solution :

    \[{{V}_{1}}=2c{{m}^{3}},\,{{V}_{2}}=?,\,h=32m\] \[{{p}_{1}}=\rho gh+{{p}_{0}},\,{{p}_{2}}={{p}_{0}}\] \[{{T}_{1}}=9+273=282\,K\] \[{{T}_{2}}=30+273=303\,K\] \[g=10m/{{s}^{2}}\] \[\rho ={{10}^{3}}kg/{{m}^{3}}\] \[\Rightarrow \] From \[\frac{{{p}_{1}}{{V}_{1}}}{{{T}_{1}}}=\frac{{{p}_{2}}{{V}_{2}}}{{{T}_{2}}}\] [Here, we assume air as an ideal gas] \[\Rightarrow \]  \[\frac{(\rho gh+{{p}_{0}})2\times {{10}^{-6}}}{282}=\frac{{{p}_{0}}\times {{V}_{2}}}{303}\] \[\Rightarrow \] \[\frac{({{10}^{3}}\times 10\times 32+{{10}^{5}})2\times {{10}^{-6}}\times 303}{282\times {{10}^{5}}}={{V}_{2}}\] \[\Rightarrow \] \[\frac{(3.2+1)\times 2\times {{10}^{-6}}\times 303}{282}={{V}_{2}}\] \[\Rightarrow \]\[\frac{4.2\times 2\times 303}{282}\times {{10}^{-6}}={{V}_{2}}\] \[\Rightarrow \] \[9\times {{10}^{-6}}={{V}_{2}}\] \[\Rightarrow \] \[{{V}_{2}}=9\,c{{m}^{3}}\] Note, Here, there is nothing said about the surface tension, so we have assumed the water pressure equal to air pressure at a particular position.

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