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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