Question

The pressure that will be build up by a compressor in a paint-gun when a stream of liquid paint flows out with a velocity of 25 $m{s^{ - 1}}$(density of the paint is 0.8$gm - c{m^{ - 3}}$) is: (in $N{m^{ - 2}}$)
(A) $2.5 \times {10^2}$
(B) $2.5 \times {10^3}$
(C) $2.5 \times {10^5}$
(D) $5 \times {10^5}$

Answer 1

Hint We should know that the Bernoulli’s Theorem states that when a horizontal flow is maintained by the liquid, then the points of the higher speed will have less pressure as compared to the ones which are at a slower fluid speed. Using this concept we can solve the following question.

Complete step by step answer
We have to apply Bernoulli's Theorem just on the inside and the outside of the holes from which the point is flowing. So, the expression becomes:
${p_1} + 0 = \dfrac{1}{2}\rho {v^2}$
$\Rightarrow {p_1} = \dfrac{1}{2} \times 800 \times {(25)^2}$
$\Rightarrow {p_1} = 2.5 \times {10^5}N/{m^2}$

Hence the correct answer is Option C.

Note We should know that the Bernoulli’s Theorem gives us an idea that the sum of the pressure energy, the kinetic energy and the potential energy per unit mass of an incompressible, non- viscous fluid in a streamlined flow will always remain as a constant.