b
(b)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 \(\frac{1}{2}{v^2} = \frac{{{P_1} - {P_2}}}{\rho }\)\( \Rightarrow \frac{1}{2}{v^2} = \frac{{3.5 \times {{10}^5} - 3 \times {{10}^5}}}{{{{10}^3}}} \Rightarrow {v^2}\)
\( = \frac{{2 \times 0.5 \times {{10}^5}}}{{{{10}^3}}} \Rightarrow {v^2} = 100 \Rightarrow v = 10\;m/s\)