A) this part can be done on the basis of equation of continuity

AV = constant

A is the area of the cross section \(\displaystyle={d}^{{2}}\)

V is the velocity of the water

\(\displaystyle{{d}_{{1}}^{{2}}}\times{V}_{{1}}={{d}_{{2}}^{{2}}}\times{V}_{{2}}\)

\(\displaystyle{d}_{{1}}={d},{d}_{{2}}={\frac{{{d}}}{{{4}}}};{V}_{{1}}={1}\ \frac{{m}}{{s}};{V}_{{2}}=?\)

plug the values and get the answer for velocity at the other end

B) We can do this part by making use of bernoullis theorem

\(\displaystyle{P}+{\frac{{{1}}}{{{2}}}}{\left({{v}_{{1}}^{{2}}}\right)}+{h}{g}={P}_{{2}}+{\frac{{{1}}}{{{2}}}}{\left({{v}_{{2}}^{{2}}}\right)}+{h}{g}\)

remains constant on both sides

therefore, \(\displaystyle{P}_{{1}}+{\frac{{{1}}}{{{2}}}}{\left({{v}_{{1}}^{{2}}}\right)}={P}_{{2}}+{\frac{{{1}}}{{{2}}}}{\left({{v}_{{2}}^{{2}}}\right)}\)

\(\displaystyle{P}_{{1}}={3.00}\times{10}^{{5}}\) Pa; \(\displaystyle{P}_{{2}}=?;{V}_{{1}}={1}\ \frac{{m}}{{s}};{V}_{{2}}=\) (get it from A);

\(\displaystyle ={1000}{k}\frac{{g}}{{m}^{{3}}}\)

plug the vales and get the answer