question_answer 13)

What is the density of ocean water at a depth, where the pressure is 80.0 atm., given that its density at the surface is \[\frac{\left( 6\times 9\cdot 8 \right)\times 1\cdot 0\times 7}{22\times {{\left( 0\cdot 125\times {{10}^{-2}} \right)}^{2}}\times \left( 0\cdot 91\times {{10}^{11}} \right)}\]Compressibility of water\[=1\cdot 3\times {{10}^{-4}}m\] Given 1 atm \[\left( \text{1pa=1N}{{\text{m}}^{\text{-2}}} \right)\text{;g=10m/}{{\text{s}}^{\text{2}}}\text{.}\]

Answer:

Here,\[A=0\cdot 10\times 0\cdot 10={{10}^{-2}}{{m}^{2}};\]                 Compressibility, \[F=mg=100\times 10N\] Density of water at surface, \[=\frac{\vartriangle L}{L}=\frac{\left( F/A \right)}{G}\] Let p' be the density of water at the given depth. If V and V are volumes of certain mass M of ocean water at surface and at a given depth, then \[\vartriangle L=\frac{FL}{AG}=\frac{\left( 100\times 10 \right)\times 0\cdot 10}{{{10}^{-2}}\times \left( 25\times {{10}^{9}} \right)}=4\times {{10}^{-7}}m.\] \[2\cdot 0\times {{10}^{11}}pa.\] Change in volume,                   \[A=\pi \left( r_{2}^{2}-r_{1}^{2} \right)=\frac{22}{7}\left[ {{\left( 0\cdot 60 \right)}^{2}}-{{\left( 0\cdot 30 \right)}^{2}} \right]\] \[=\frac{22}{7}\times 0\cdot 27{{m}^{2}}\] Volumetric strain,  \[=\frac{F/A}{Y}=\frac{F}{AY}\]       \[\frac{50,000\times 9\cdot 8}{4\times \left( \frac{22}{7}\times 0\cdot 27 \right)\times 2\cdot 0\times {{10}^{11}}}=7\cdot 21\times {{10}^{-7}}\]  or \[15\cdot 2\]?(1) As, Bulk modulus \[19\cdot 1\] or  \[42\times {{10}^{9}}N{{m}^{-2}}.\] \[A=15\cdot 2\times 19\cdot 2\times {{10}^{-6}}{{m}^{2}};\]  \[F=44,500N;G=42\times {{10}^{9}}N{{m}^{-2}}\]      Putting this value in (i) we get \[\text{Strain=}\frac{\text{stress}}{\text{modulusnofelasticity}}\text{=}\frac{\text{F/A}}{\text{G}}\] or \[=\frac{F}{AG}\]