Water flows through a horizontal tube as shown in figure. If the difference of heights of water column in the vertical tubes is 2 cm, and the areas of cross section at A and B are 4 cm^{2} and 2 cm^{2} respectively, find the rate of flow of water across any section.

#### Solution

Given:

Difference in the heights of water columns in vertical tubes = 2 cm

Area of cross section at A, a_{A} = 4 cm^{2}

Area of cross section at B, a_{B} = 2 cm^{2}

Now, let v_{A}_{ }and v_{B}_{ }be the speeds of water at A and B, respectively.

From the equation of continuity, we have:

\[v_A a_A = v_B \times a_B \]

\[ \Rightarrow v_A \times 4 = v_B \times 2\]

\[ \Rightarrow v_B = 2 v_A . . . (i)\]

From Bernoulli's equation, we have:

\[\frac{1}{2}\rho v_A^2 + ρg h_A + p_A = \frac{1}{2}\rho v_B^2 +ρg h_B + p_B \]

\[ \Rightarrow \frac{1}{2}\rho v_A^2 + p_A = \frac{1}{2}\rho v_B^2 + p_B \]

\[ \Rightarrow p_A - p_B = \frac{1}{2}\rho\left( v_B^2 - v_A^2 \right)\]

Here,

p_{A} and p_{B} are the pressures at A and B, respectively.

h_{A} and h_{B} are the heights of water columns at point A and B, respectively.

ρ is the density of the liquid.

Thus, we have:

\[\frac{1}{2} \times 1 \times \left( 4 v_A^2 - v_A^2 \right)\]

\[ \Rightarrow 2 \times 1 \times 1000 = \frac{1}{2} \times 1 \times 3 v_A^2 \]

\[[ p_A - p_B = 2 cm = 2 \times 1 \times 1000 {\text{ dyne }/ {\text{cm}}^2} (\text{water column})]\]

\[ \Rightarrow v_A^2 = \sqrt{\frac{4000}{3}} = 36 . 51 cm/s\]

\[ \therefore\text{ Rate of flow }= v_A a_A = 36 . 51 \times 4\]

\[ = 146 {\text{ cm}}^3 /s\]

Hence, the required rate of flow of water across any section is 146 cm^{3}/s.