At the bottom of tank pressure is \(3\) atmosphere. So, total pressure due to water column

\( = h\rho g = 2 \times {10^5}\,\left( {two\,atmosphere} \right)\)

\( \Rightarrow \,gh = \frac{{2 \times {{10}^5}}}{\rho } = \frac{{2 \times {{10}^5}}}{{{{10}^3}}} = 2 \times {10^2}\)

\( \Rightarrow v = \sqrt {2 \times 2 \times {{10}^2}}  = \sqrt {400} \,m/\sec \)