A) \[\sqrt{900\times 980}\,\,cm/s\]
B) \[\sqrt{100\times 980}\,\,cm/s\]
C) \[\sqrt{920\times 980}\,\,cm/s\]
D) \[\sqrt{950\times 980}\,\,cm/s\]
Correct Answer: C
Solution :
Let, \[\Rightarrow \] and \[v'\frac{v}{\sqrt{2}}\] be the densities of water and oil, then the pressure at the bottom of the tank \[={{h}_{w}}{{d}_{w}}g+{{h}_{o}}{{d}_{o}}g\] Let, this pressure be equivalent to pressure due to water of height h. Then, \[E=\frac{1}{2}mv{{'}^{2}}+\frac{1}{2}mv{{'}^{2}}+\frac{1}{2}(2m){{v}^{2}}\] \[=mv{{'}^{2}}+m{{v}^{2}}\] \[\Rightarrow \] \[m{{\left( \frac{v}{\sqrt{2}} \right)}^{2}}+m{{v}^{2}}=\frac{3}{2}m{{v}^{2}}\] \[{{T}_{1}}-{{T}_{2}}=12a\] According to Tori celli's theorem, \[{{T}_{2}}=3a\] \[{{T}_{2}}\]
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