A) 50\[\frac{{{\varepsilon }_{0}}A}{d}\left[ \frac{{{k}_{1}}}{{{k}_{2}}}+\frac{({{k}_{2}}+{{k}_{3}})}{{{k}_{2}}{{k}_{3}}} \right]\]
B) 51 \[\frac{{{\varepsilon }_{0}}A}{d}\left[ \frac{2}{{{k}_{1}}}+\frac{{{k}_{2}}{{k}_{3}}}{{{k}_{2}}+{{k}_{3}}} \right]\]
C) 40.2 \[\frac{{{\varepsilon }_{0}}A}{d}\left[ \frac{2}{{{k}_{1}}}+\frac{{{k}_{2}}+{{k}_{3}}}{{{k}_{2}}{{k}_{3}}} \right]\]
D) 60m\[1.81\times {{10}^{-5}}wb\]
Correct Answer: A
Solution :
Let a be the area of hole and \[{{v}_{e}}\] be the efflux velocity from the hole at a depth h below the liquid level. Let A be area of cross-section of tank and v be the speed with which level decreases in the container. \[\therefore \] By equation of continuity, \[a{{v}_{e}}=Av\] \[v=\frac{a{{v}_{e}}}{A}\] By Bernoullis theorem \[{{p}_{0}}+h\rho g+\frac{1}{2}\rho {{v}^{2}}={{p}_{0}}+\frac{1}{2}pv_{e}^{2}\] \[h\rho g+\frac{1}{2}\rho .\frac{{{a}^{2}}v_{e}^{2}}{{{A}^{2}}}=\frac{1}{2}\rho v_{e}^{2}\] \[v_{e}^{2}=\frac{2gh}{1-\frac{{{a}^{2}}}{{{A}^{2}}}}=\frac{2\times (3-0.525)\times 10}{1-{{(0.1)}^{2}}}\] \[=50{{m}^{2}}{{s}^{-2}}\]
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