A) \[5.937\text{ }c{{m}^{3}}\]
B) \[8.937\text{ }c{{m}^{3}}\]
C) \[12.937\text{ }c{{m}^{3}}\]
D) \[16.937\text{ }c{{m}^{3}}\]
Correct Answer: B
Solution :
\[{{V}_{1}}=2\,c{{m}^{3}},{{V}_{2}}=?h=32\,m\] \[{{\rho }_{1}}=\rho gh+{{\rho }_{0}},{{\rho }_{2}}={{\rho }_{0}}\] \[{{T}_{1}}=9+273=282\,K\] \[{{T}_{2}}=30+273=303\,K\] \[g=10\,m/{{s}^{2}}\] \[\rho ={{10}^{3}}\,kg/{{m}^{3}}\] \[\Rightarrow \] Form\[\frac{{{p}_{1}}{{V}_{1}}}{{{T}_{1}}}=\frac{{{\rho }_{2}}{{V}_{2}}}{{{T}_{2}}}\] [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.
You need to login to perform this action.
You will be redirected in
3 sec
