• # 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)  $16.937\,\,c{{m}^{3}}$

${{V}_{1}}=2c{{m}^{3}},\,{{V}_{2}}=?,\,h=32m$ ${{p}_{1}}=\rho gh+{{p}_{0}},\,{{p}_{2}}={{p}_{0}}$ ${{T}_{1}}=9+273=282\,K$ ${{T}_{2}}=30+273=303\,K$ $g=10m/{{s}^{2}}$ $\rho ={{10}^{3}}kg/{{m}^{3}}$ $\Rightarrow$ 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.