question_answer 12)

Compute the bulk modulus of water from the following data: Initial volume \[{{Y}_{2}}\frac{{{F}_{2}}\times {{l}_{2}}}{{{a}_{2}}\times \vartriangle {{l}_{2}}}=\frac{F}{3\cdot 0\times {{10}^{-5}}}\times \frac{4\cdot 7}{\vartriangle l}\] litre, pressure increase = 100-0 atmosphere. Final volume \[\therefore \] litre. (1 atmosphere \[\frac{{{Y}_{1}}}{{{Y}_{2}}}=\frac{4\cdot 7\times 4\times {{10}^{-5}}}{3\cdot 5\times 3\cdot 0\times {{10}^{-5}}}=1\cdot 8\]). Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.

Answer:

Here, \[{{Y}_{1}}:{{Y}_{2}}=1\cdot 8:1\]litre \[=150\times {{10}^{6}}N{{m}^{-2}},\] \[\text{Y=}\frac{\text{Stress}}{\text{Strain}}\text{=}\frac{\text{150 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{6}}}}{\text{0 }\!\!\times\!\!\text{ 002}}\] \[=7\cdot 5\times {{10}^{10}}N{{m}^{-2}}\]liter or\[=300\times {{10}^{6}}N{{m}^{-2}}\] \[=3\times {{10}^{8}}N{{m}^{-2}}\]\[\begin{align} & {{F}_{1}}=4+6=10kgf=10\times 9\cdot 8N; \\ & {{l}_{1}}=1\cdot 5m,\vartriangle {{l}_{1}}=?;2{{r}_{1}}=0\cdot 25cm \\ \end{align}\] litre \[\begin{align} & {{r}_{1}}=\left( 0\cdot 25/2 \right)cm=0\cdot 125\times {{10}^{-2}}m, \\ & {{Y}_{1}}=2\cdot 0\times {{10}^{11}}pa \\ \end{align}\] We know that bulk moduls, \[{{\text{F}}_{\text{2}}}\text{=6 }\!\!\times\!\!\text{ 0kgf=6 }\!\!\times\!\!\text{ 9 }\!\!\times\!\!\text{ 8N;2}{{\text{r}}_{\text{2}}}\text{=0 }\!\!\times\!\!\text{ 25cm}\] \[\begin{align} & {{r}_{2}}=\left( 0\cdot 25/2 \right)cm=0\cdot 125\times {{10}^{-2}}m; \\ & {{Y}_{2}}=0\cdot 91\times {{10}^{11}}pa,{{l}_{2}}=1\cdot 0m,\vartriangle {{l}_{2}}=? \\ \end{align}\] \[{{Y}_{1}}=\frac{{{F}_{1}}\times {{l}_{1}}}{{{a}_{1}}\times \vartriangle {{l}_{1}}}=\frac{{{F}_{1}}\times {{l}_{1}}}{\pi r_{1}^{2}\times \vartriangle {{l}_{1}}}\] Bulk modulus of air \[\vartriangle {{l}_{1}}=\frac{{{F}_{1}}\times {{l}_{1}}}{\pi r_{1}^{2}\times {{Y}_{1}}}\] \[\frac{\left( \text{10 }\!\!\times\!\!\text{ 9 }\!\!\times\!\!\text{ 8} \right)\text{ }\!\!\times\!\!\text{ 1 }\!\!\times\!\!\text{ 5 }\!\!\times\!\!\text{ 7}}{\text{22 }\!\!\times\!\!\text{ }{{\left( \text{0 }\!\!\times\!\!\text{ 125 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-2}}} \right)}^{\text{2}}}\text{ }\!\!\times\!\!\text{ 2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{11}}}}\] \[=1\cdot 49\times {{10}^{-4}}m\] \[\vartriangle {{l}_{2}}=\frac{{{F}_{2}}\times {{l}_{2}}}{\pi r_{2}^{2}\times {{Y}_{2}}}=\] It is so because gases are much more compressible than those of liquids. The molecules in gases are very poorly coupled to their neighbours as compared to those of gases.