A) \[6.93\times {{10}^{-4}}{{m}^{3}}{{s}^{-1}}\]
B) \[7.8\times {{10}^{-4}}{{m}^{3}}{{s}^{-1}}\]
C) \[10.4\times {{10}^{-5}}{{m}^{3}}{{s}^{-1}}\]
D) \[14.5\times {{10}^{-5}}{{m}^{3}}{{s}^{-1}}\]
Correct Answer: A
Solution :
From Bernoullis theorem \[{{p}_{1}}+\frac{1}{2}\rho v_{1}^{2}={{p}_{2}}+\frac{1}{\alpha }\rho v_{2}^{2}\] \[\therefore \] \[{{p}_{1}}-{{p}_{2}}=\frac{1}{2}\rho (v_{2}^{2}-v_{1}^{2})\] \[\therefore \] \[10=\frac{1}{2}\times 1.25\times {{10}^{3}}(v_{2}^{2}-v_{1}^{2})\] \[\therefore \] \[v_{2}^{2}-v_{1}^{2}=\frac{10\times 2}{1.25\times {{10}^{3}}}=16\times {{10}^{-3}}\] ... (i) Also from equation of continuity \[={{A}_{1}}{{v}_{1}}={{A}_{2}}{{v}_{2}}\] \[\pi r_{1}^{2}{{v}_{1}}=\pi r_{2}^{2}{{v}_{2}}\] \[\therefore \] \[\frac{{{v}_{1}}}{{{v}_{2}}}={{\left[ \frac{{{r}_{2}}}{{{r}_{1}}} \right]}^{2}}=\frac{0.04}{0.1}=0.4\] \[{{v}_{1}}=0.4{{v}_{2}}\] ... (ii) Substituting this value in Eq. (i) \[v_{2}^{2}-{{(0.4{{v}_{2}})}^{2}}=16\times {{10}^{-3}}\] \[{{v}_{2}}=1.38\times {{10}^{-1}}=0.138m{{s}^{-1}}\] Rate of flow of glycrine\[v={{A}_{2}}{{v}_{2}}\] \[=\pi r_{2}^{2}{{v}_{2}}\] \[=6.93\times {{10}^{-4}}\times {{m}^{3}}{{s}^{-1}}\]
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