Question

The flow of water in a horizontal pipe is streamline. At a point in the tube where its cross-sectional area is $ 10c{m^2} $ , the velocity of water and pressure are $ 1m/s $ and $ 2 \times {10^3} $ pascal, respectively. The pressure of water at a different point, where the cross-sectional area is $ 5c{m^2} $ will be:
A) 2000 Pascal
B) 1000 Pascal
C) 250 Pascal
D) 500 Pascal

Answer 1

Hint: In this solution, we will use the continuity equation to determine the velocities where the cross-sectional area is $ 5c{m^2} $ . And then we will use Bernoulli's Equation to solve for the pressure at that point

Formula used: In this solution, we will use the following formulae:
-Continuity equation: $ {A_1}{v_1} = {A_2}{v_2} $ where $ A $ is the area of the tube at a particular point and $ v $ is the velocity of water at that point in the tube. Different positions in the tube are denoted by subscripts 1 and 2.
- Bernoulli’s equation: $ {P_{atm}} + \rho gh + \dfrac{1}{2}\rho {v^2} = {\text{P}} $ where $ {P_{atm}} $ is the atmospheric pressure, $ \rho $ is the density of the liquid, $ g $ is the gravitational acceleration, $ h $ is the height of the liquid above our reference point and $ v $ is the velocity of water, and $ P $ is the total pressure.

Complete step by step answer:
To solve this question, we start by using the continuity equation which relates the areas and velocities of the tube and the liquid in different parts of the pipe. Writing the continuity equation for the two points of the pipe, we can write
 $ {A_1}{v_1} = {A_2}{v_2} $
Substituting $ {A_1} = 10c{m^2} $ , $ {v_1} = \,1m/s $ , $ {A_2} = 5\,c{m^2} $ , we get
 $ {v_2} = 2\,m/s $
Since the pipe is horizontal, there is no difference in the height of the water or its density. So, using Bernoulli’s theorem we can write at the two points of the pipe,
 $ {P_2} - {P_1} = \dfrac{1}{2}\rho (v_2^2 - v_1^2) $
Substituting the values of the velocity at the two points, and the density of water as $ 1000\,kg/{m^3} $ , we get
 $ {P_2} - {P_1} = \dfrac{1}{2} \times 1000 \times \left( {{1^2} - {2^2}} \right) $
Which gives us
 $ {P_2} - 2000 = - 1500 $
Or alternatively,
 $ {P_2} = 500Pa $
Which will be the pressure at the second point of the pipe.
So the correct answer will be option D.

Note:
The first term on the left side of Bernoulli’s theorem is due to external pressure, the second is due to pressure due to depth and the third is due to velocity of flow of water. In our case, the pressure difference will only arise due to the difference of velocities of the flow of water which we should remember to determine through the continuity theorem.