Water is flowing through a horizontal tube as depicted in the...

Given, area of cross-section at $A$, $A_1 = 4 \text{ cm}^2$

Area of cross-section at $B$, $A_2 = 2 \text{ cm}^2$

Difference in height of water columns, $h = 2 \text{ cm}$

Let $v_1$ and $v_2$ be the velocities of water at $A$ and $B$ respectively.

From the equation of continuity:

$A_1 v_1 = A_2 v_2$

$4 v_1 = 2 v_2 \Rightarrow v_2 = 2 v_1$

Applying Bernoulli's theorem for the horizontal tube:

$P_1 + \dfrac{1}{2} \rho v_1^2 = P_2 + \dfrac{1}{2} \rho v_2^2$

$P_1 - P_2 = \dfrac{1}{2} \rho (v_2^2 - v_1^2)$

The pressure difference is also given by the hydrostatic pressure of the water column difference:

$P_1 - P_2 = \rho g h$

Equating the two expressions for pressure difference:

$\rho g h = \dfrac{1}{2} \rho (v_2^2 - v_1^2)$

$2 g h = (2 v_1)^2 - v_1^2 = 3 v_1^2$

$v_1 = \sqrt{\dfrac{2 g h}{3}}$

Taking $g = 1000 \text{ cm/s}^2$:

$v_1 = \sqrt{\dfrac{2 \times 1000 \times 2}{3}} = \sqrt{\dfrac{4000}{3}} \text{ cm/s}$

The rate of flow of water $Q$ is:

$Q = A_1 v_1 = 4 \times \sqrt{\dfrac{4000}{3}}$

$Q \approx 4 \times 36.51 = 146.04 \text{ cm}^3/\text{s} \approx 146 \text{ cm}^3/\text{s}$

Answer: $146 \text{ cm}^3/\text{s}$