Given that, pressure difference


By using Bernoulli's equation,
$$
p_A+\frac{1}{2} \rho v_A^2=p_B+\frac{1}{2} \rho v_B^2
$$

Density of water, $\rho=10^3 \mathrm{~kg} \mathrm{~m}^{-3}$
Cross-section areas at points $A$ and $B$,
$$
\begin{aligned}
& a_A=40 \mathrm{~cm}^2=40 \times 10^{-4} \mathrm{~m}^2 \\
& a_B=20 \mathrm{~cm}^2=20 \times 10^{-4} \mathrm{~m}^2
\end{aligned}
$$
By equation of continuity, rate of flow of water through the tube
$$
\frac{\Delta V}{\Delta t}=a_A v_A=a_B v_B \Rightarrow \frac{v_A}{v_B}=\frac{a_B}{a_A}=\frac{1}{2}
$$

Substituting the values from Eqs. (i) and (iii), in Eq. (ii), we get
$$
\begin{aligned}
1500 & =\frac{1}{2} \times 10^3\left[\left(2 v_A\right)^2-v_A^2\right] \\
1500 & =500\left(4 v_A^2-v_A^2\right) \\
v_A^2 & =1 \Rightarrow v_A=1 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
$\therefore$ Rate of flow of water,
$$
\begin{aligned}
\frac{\Delta V}{\Delta t} & =a_A v_A=40 \times 10^{-4} \times 1 \mathrm{~m}^3 / \mathrm{s} \\
& =4000 \mathrm{~cm}^3 / \mathrm{s}
\end{aligned}
$$