A) \[\rho ={{\rho }_{0}}\left[ 1-\frac{{{\rho }_{0}}gy}{B} \right]\]
B) \[\rho ={{\rho }_{0}}\left[ 1+\frac{{{\rho }_{0}}gy}{B} \right]\]
C) \[\rho ={{\rho }_{0}}\left[ 1+\frac{B}{{{\rho }_{0}}hgy} \right]\]
D) \[\rho ={{\rho }_{0}}\left[ 1-\frac{B}{{{\rho }_{0}}hgy} \right]\]
Correct Answer: B
Solution :
Bulk modulus, \[B=-{{V}_{0}}\frac{\Delta p}{\Delta V}\,\,\Rightarrow \,\Delta V=-{{V}_{0}}\frac{\Delta p}{B}\] \[\Rightarrow \,\,\,\,V={{V}_{0}}\left[ 1-\frac{\Delta p}{B} \right]\] \[\therefore \,\,\,\,Density,\,\,\,\rho ={{\rho }_{0}}\,{{\left[ 1-\frac{\Delta p}{B} \right]}^{-1}}\,\,=\,\,{{\rho }_{0}}\left[ 1+\frac{\Delta p}{B} \right]\] where, \[\Delta p=p\,-{{p}_{0}}=\,\,h{{\rho }_{0}}g\] = pressure difference between depth and surface of ocean \[\therefore \,\,\,\,\,\rho ={{\rho }_{0}}\left[ 1+\frac{{{\rho }_{0}}gy}{B} \right]\,\,\,\,\,(As\,\,h=y)\]
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