At a point in a pipeline the velocity of liquid is 2 m/s and the pressure is 10^5 Pa. what is the pressure at a point 10 m lower than the first one (the cross-section is 1/4th of the previous one and the density of liquid is 800 kg/m^3)

To find the pressure at a point 10 m lower in the pipeline, we can use the principles of fluid mechanics and Bernoulli's equation.

Bernoulli's equation states that the total energy per unit volume of a fluid remains constant along a streamline. It can be expressed as:

P1 + 1/2ρv1^2 + ρgh1 = P2 + 1/2ρv2^2 + ρgh2

where:
P1 and P2 are the pressures at the two points
ρ is the density of the liquid
v1 and v2 are the velocities of the liquid at the two points
g is the acceleration due to gravity
h1 and h2 are the heights of the two points above a reference level

In this case, the two points are at the same height above the reference level, so we can ignore the terms involving h1 and h2.

Given:
v1 = 2 m/s
P1 = 10^5 Pa
ρ = 800 kg/m^3
The cross-sectional area is 1/4th of the previous one, which means the velocity at point 2 will be 4 times the velocity at point 1.

Using Bernoulli's equation, we can solve for P2:

P2 = P1 + 1/2ρ(v1^2 - v2^2)

P2 = 10^5 + 1/2 * 800 * (2^2 - 4^2)

P2 = 10^5 - 1/2 * 800 * (-12)

P2 = 10^5 + 1/2 * 800 * 12

P2 = 10^5 + 4800

P2 = 10^5 + 4.8 × 10^3

P2 = 10^5 + 4.8 × 10^3

P2 = 10^5 + 4800

P2 = 104800 Pa

Therefore, the pressure at a point 10 m lower in the pipeline is approximately 104800 Pa.