A horizontal pipeline carries water in streamline flow. At a point along the pipe, where the cross-sectional area is $10\,c{m^2}$, the water velocity is $1\,m{s^{ - 1}}$ and the Pressure is $2000\,Pa$. The Pressure of water at another point where the cross-sectional area is $5\,c{m^2}$is _______ Pa (Density of water = ${10^3}\,kg\,{m^{ - 3}}$)

Hint – Here, we first need to find out volume\[\,\,\left( {{V_2}} \right)\]using continuity equation and then by using Bernoulli’s equation pressure at the other point can also be detected, as all the other values are already given. Using this one can easily find the solution to the given problem.

Step-By-Step answer:
Let’s first state all the given values -
$ \Rightarrow {A_1}\, = \,10\,c{m^2},\,\,{V_1}\, = \,1\,m/s,\,\,{P_1}\, = \,2000\,Pa$
\[ \Rightarrow {A_2}\, = \,5\,c{m^2},\,\,{V_2}\, = ?,\,{P_2}\, = \,?,\,Density\, = \,{10^3}\]
Where, ${A_1}$ stands for the cross-sectional area at given point , ${V_1}$ stands for the water velocity, ${P_1}$ is the water pressure at the above mentioned cross section, ${A_2}$ is the cross sectional area at another point, ${V_2}$ is the velocity of water, ${P_2}$ is the pressure at cross section ${A_2}$and density of water is ${10^3}\,kg\,{m^{ - 3}}$.
Continuity principle can be defined as the principle of fluid mechanics. In simple terms, what goes into a fixed volume in a given time, minus what flows out of the said volume in that time, should add up to the original volume i.e. it should accumulate in that volume.
Now, according to the continuity equation,
$ \Rightarrow {A_1}{V_1}\, = \,{A_2}{V_2}$
$ \Rightarrow \,\,{V_2}\, = \,\dfrac{{{A_1}{V_1}}}{{{A_2}}}$
$i.e.\,\,{V_2}\, = \,\,\dfrac{{10 \times 1}}{5}\,\, = \,\,2\,m/s$ - equation (1)

Bernoulli’s Equation –
Bernoulli’s equation states that when there is a decrease in the static pressure or the potential energy of the fluid, it results in the increase of the Speed of the fluid.
i.e. Speed of the fluid is inversely proportional to the static pressure / potential energy
This principle can be used when the behaviour of the fluid flow is known (as in this case it is clearly mentioned that the pipeline carries water in streamline flow)

Now for finding the pressure at the other point $P_2$, we use Bernoulli’s equation, substituting the value of ${V_2}$ from the equation (1),
$ \Rightarrow $ Bernoulli’s Equation = ${P_1}\, + \,\dfrac{1}{2}\rho\,{V_1}^2\, = \,{P_2}\, + \,\dfrac{1}{2}\rho\,{V_2}^2$
$ \Rightarrow \,\,2000\, + \,\dfrac{1}{2} \times \,{10^3}\, \times \,1\, = \,x\, + \,\dfrac{1}{2}\, \times \,{10^3}\, \times \,2$
$
   \Rightarrow \,2500\, = \,x\, + \,2000 \\
   \Rightarrow \,x\, = \,2500\, - \,2000\, = \,500\,Pa \\
 $
i.e. ${P_2}\, = \,500\,Pa$
Therefore, the pressure at other point is 500Pa

Note – It is important to have a clear knowledge of the continuity equation and Bernoulli’s equation in order to solve problems of this kind. Continuity equation is used when one of the quantities is unknown and it obeys the Law of conservation, thus through this equation water velocity can be identified. Bernoulli’s equation is a relation pressure and can be used when the behaviour of the fluid flow is known.