Question

The figure shows a crude type of perfume atomizer. When the bulb at A is compressed, air flows swiftly through tiny BC with uniform speed V, thereby causing a reduced pressure at the position of the vertical tube DE. The liquid density is $500kg/{{m}^{3}}$, then rises in the tube, enters tube BC and sprayed out. When the bulb is in its natural position the air in the bulb and tube are at atmospheric pressure ${{P}_{0}}^{5}=15N/{{m}^{2}}$ . When bulb A is compressed.it creates an excess pressure $\Delta \rho =0.001{{P}_{0}}$ inside the bulb A. density of air is $1.3kg/{{m}^{3}}$ . If the magnitude of minimum value of speed V. required to cause the liquid to rise to tube BC is $5Km/s$ . Find the value of k, ($g=10m/{{s}^{2}}$ ).

Answer 1

Hint: Pressure is the amount of force applied normal to the surface area of an object.it is also stated as the force applied per unit area which is perpendicular to the surface. Pressure is designed with the letter P and can also be used on some occasions. PSI is the unit of pressure or stress based on avoirdupois units. A pound per square inch which clears that pressure is relative to a vacuum rather than the ambient atmospheric pressure.

Complete step-by-step answer:
Velocity is defined as the rate of change of displacement with respect to time and in kinematics velocity is a fundamental concept.SI unit of velocity is $m{{s}^{-1}}$ velocity tracking is the measure of velocity.
Velocity (v) =$\dfrac{\Delta S}{\Delta t}$
The dimensional formula of velocity is ${{M}^{0}}{{L}^{1}}{{T}^{-1}}$.
By making use of Bernoulli’s equation at point 3 and point 4, then we can write the below equation
$\dfrac{{{P}_{0}}+\Delta P}{{{\rho}_{liquid}}}+\dfrac{1}{2}{{v}^{2}}_{liquid}={{P}_{2}}^{'}\times{{\rho }_{liquid}}$
On comparing this equation with the question
\[\begin{align}
  &\dfrac{{{P}_{BC}}+{{\rho}_{liquid}}gh}{{{\rho}_{liquid}}}+\dfrac{1}{2}{{v}_{1}}^{2}=\dfrac{{{P}_{0}}}{{{\rho }_{1}}} \\
 &\Rightarrow\dfrac{{{P}_{0}}+\Delta P+{{\rho}_{1}}gh-0.65{{v}^{2}}}{{{\rho}_{1}}}
+ \dfrac{1}{2}{{v}^{2}}_{1} \\
 & \dfrac{1}{2}{{\rho }_{1}}{{v}_{1}}^{2}=0.65{{v}^{2}}-\Delta P-{{\rho }_{1}}gh\ge 0 \\
 & \Rightarrow {{v}^{2}}\ge \dfrac{\Delta P+{{\rho }_{1}}gh}{0.65} \\
 & {{v}_{\min }}=\sqrt{\dfrac{\Delta P+{{\rho }_{1}}gh}{0.65}}=\dfrac{100}{5}=2 \\
\end{align}\]

Note: The atmosphere ($atm$ ) is another unit of pressure widely used.1atm of pressure is considered the normal ambient pressure which is equivalent to $760mmHg$ which is$101.3kPa$. Pounds square inch (psi) is often also specified at atmosphere pressure. Blaise Pascal was also an early pioneer of “The game theory”.