Question

Water flows through a tube shown in figure. The area of cross section at A and B are 1 cm² and 0.5 cm² respectively. The height difference between A and B is 5 cm. If the speed of water at A is 10 cm s find (a) the speed at B and (b) the difference in pressures at A and B.

Answer 1

Given:
Difference in the heights of A and B = 5 cm
Area of cross section at A, a_a = 1 cm²
Area of cross section at B, a_b = 0.5 cm²
Speed of water at A, \[v_A\]=10 cms

(a) From the equation of continuity, we have:

\[ \vec{V}_A \times a_A = \vec{V}_B \times a_B \]

\[ \Rightarrow 10 \times 1 = \vec{V}_B \times 0 . 5\]

\[ \Rightarrow \vec{V}_B = 20 \text{ cm}/s\]

The required speed of water at cross section B is 20 cms

(b) From Bernoulli's equation, we get:

\[ \frac{1}{2}\rho v_A^2 + ρ g h_A + P_A = \frac{1}{2}\rho v_B^2 + ρ g h_A + P_B \]

\[ \Rightarrow P_B - P_A = \frac{1}{2}\rho\left( v_A^2 - v_B^2 \right) + ρg\left( h_A - h_B \right)\]

Here,
P_A and P_B are the pressures at A and B, respectively.
h_A and h_B are the heights of points A and B, respectively.
ρ is the density of the liquid.
On substituting the values, we have:

\[P_B - P_A = \frac{1}{2} \times 1(100 - 400) + 1 \times 1000(5 . 0)\]

\[ = - 150 + 5000 = 4850 \text{ Dyne}/ {cm}^2 \]

\[ = 485 N/ m^2\]

herefore, the required pressure difference at A and B is 485 N/m².