A) 100 m/s
B) 10 m/s
C) 1 m/s
D) \[10\sqrt{10}\] m/s
Correct Answer: B
Solution :
Bernoulli's theorem for unit mass of liquid \[\frac{P}{\rho }+\frac{1}{2}{{v}^{2}}=\] constant As the liquid starts flowing, it pressure energy decreases \[\frac{1}{2}{{v}^{2}}=\frac{{{P}_{1}}-{{P}_{2}}}{\rho }\]\[\Rightarrow \frac{1}{2}{{v}^{2}}=\frac{3.5\times {{10}^{5}}-3\times {{10}^{5}}}{{{10}^{3}}}\Rightarrow {{v}^{2}}\] \[=\frac{2\times 0.5\times {{10}^{5}}}{{{10}^{3}}}\Rightarrow {{v}^{2}}=100\Rightarrow v=10\ m/s\]
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