Explanation

(a) Bernoulli’s equation for flowing liquid be written as
$p + \frac{1}{2}{\rho v}^{2} + \rho gh = constant$ …(i)
Dividing the Eq. (i)by $\rho g$, we have
$\frac{P}{\rho g} + \frac{v^{2}}{2g} + h = constant$
$\ $In this expression$,\frac{v^{2}}{2g}$ is velocity head and$\ \frac{P}{\rho g}\ $is pressure head.
It is given that,
$Velocity\ head = pressure\ head$
$ie,\ \ \ \ \frac{v^{2}}{2g} = \frac{P}{\rho g}$
$or\ \ \ \ \ v^{2} = \frac{2p}{\rho}$
$or\ = \frac{2 \times 13.6 \times 10^{3} \times 40 \times 10^{- 2} \times 9.8}{10^{3}}$
$\therefore\ \ v = 10.32\ {ms}^{- 1}$