\(A _{1} v _{1}= A _{2} v _{2}\)
\(5^{2} \times 4=2^{2} \times v _{2}\)
\(v _{2}=25\)
Apply the energy equation at both the ends.
\(P _{1}+\frac{1}{2}\rho v _{1}^{2}= P _{2}+\frac{1}{2}\rho v _{2}^{2}\)
\(P_{1}-P_{2}=\frac{1}{2}\rho \left(v_{2}^{2}-v_{1}^{2}\right)\)
\(=\frac{1}{2} \times 10^{3}\left(25^{2}-4^{2}\right)\)
\(=304500 Pa\)