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This is second of our many quizzes on Fluid pressure. Click on this to access more such competitive quizzes on Pressure In Fluids.

Fluid pressure Quiz 2

physics

A manometer is a U-shaped device filled with a fluid and measures air pressure. Suppose you have a manometer filled with water, with one end of the manometer open to the atmosphere. You have a gas cylinder whose pressure you want to measure so you connect the efflux of gas from the cylinder to the other end of the U-tube. Determine the absolute pressure of the gas cylinder, if the difference in water levels between the two arms of the $U-tube$ is $5\,cm$ and the atmospheric pressure is $10332\,mm\,of\,H_2O$. $(\text{Take}\,1\,mm\,of\,H_2O = \dfrac{1}{13.6}\,mm\,of\,Hg)$ 

First determine the gauge pressure of the cylinder, following which you can arrive at its absolute pressure. Pay attention to the units used and their conversion. 

  $\text{Given that}\,h =5\,cm\,\text{and}\,P_0 = 10332\,mm\,of\,H_2O\\ \text{Let the density of water be}\,\rho = 1000\,kgm^{-3}.\\ \text{The gauge pressure of the cylinder will be:}\\ P_{gauge} = 5\,cm\,of\,H_2O\\ \Rightarrow P_{gauge}= 50mm\,of\,H_2O\\ \text{The absolute pressure of the cylinder will be:}\\ P_{abs} = P_0 + P_{gauge}\\ \Rightarrow P_{abs} = 10332 + 50\\ \Rightarrow P_{abs} = 10382\,mm\,of\,H_2O\\ \Rightarrow P_{abs} = 10382 \times \dfrac{1}{13.6}\\ \therefore P_{abs}= 763.4\,mm\,of\,Hg$

A fire hydrant is a device that collects water from pipes and water distribution systems in urban and suburban areas, enabling firefighters to rapidly access the area's water source to put out fires.You've already seen how, when a car collides with a fire hydrant and pops the spray cap off, water shoots up into the air like a geyser and spills over. This is due to the fact that the fire hydrant receives high-pressure water.If the water flows through the hydrant with a velocity of $2\,ms^{-1}$ and at a pressure of $2 \times 10^{5}\,Pa$ then determine the velocity with which the water rushes out into the air. $(\text{Take density of water}\,\rho = 1000\,kgm^{-3}\,\text{and atmospheric pressure}\,P_a = 1.013 \times 10^{5}\,Pa)$ 

Remember that the total mechanical energy associated with the motion of a fluid in a hydrodynamic system remains constant. 

 $\text{Given that density of water}\,\rho = 1000\,kgm^{-3}\\ \text{When the water flows through the hydrant:}\\ P_1 = 2 \times 10^{5}\,Pa\,\text{with velocity}\,v_1 = 2\,ms^{-1}\\ \text{When the water flows out of the hydrant:}\\ P_2 = 1.013 \times 10^{5}\,Pa\,\text{with}\,v_2\\ \text{Assuming that the height from which the hydrant expels water is constant,}\\ \text{we have the Bernoulli’s equation:}\\P_1 + \dfrac{1}{2}\rho v_1^2 = P_2 + \dfrac{1}{2}\rho v_2^2\\ \Rightarrow v_2 = \sqrt{\dfrac{2}{\rho}\left(\dfrac{1}{2}\rho v_1^2 + P_1-P_2\right)}\\ \Rightarrow v_2 = \sqrt{\dfrac{2}{1000}\left(\dfrac{1}{2} \times 1000 \times  2^2 + (2-1.013)\times 10^{5}\right)}\\ \therefore v_2 = 14.19 \approx 14.2\,ms^{-1}$

You decide to make some jelly for dessert and you pour your gelatine mix into a glass. After the gelatine has set, you notice that an air bubble is trapped in the middle of the jelly. If the bubble experiences a force $F_1$ due to an external pressure $P_1$ and force $F_2$ due to internal pressure $P_2$, find the radius $\,R$ of the bubble if it experiences a surface tension of $\,T$.  

List the forces that hold the air bubble in equilibrium and use an expression to measure the radius of the bubble.

 $\text{Let the radius of the bubble be R.}\\ \text{Let us consider a hemispherical portion of the air bubble and the forces acting on it.}\\ \text{The force due to surface tension T acting inwards from the rim of the bubble will be:}\\ F_{T} = 2\pi R T\\ \text{The force}\,F_1\,\text{due to an external pressure}\,P_1\,\text{acting radially inward}\\ \text{across the cross-sectional area of the bubble will be:}\\ F_1 = P_1A = P_1(\pi R^2)\\ \text{The force}\,F_2\,\text{due to internal pressure}\,P_2\,\text{acting radially outward}\\ \text{across the cross-sectional area of the bubble will be:}\\ F_2 = P_2A = P_2(\pi R^2)\\ \text{Since the air bubble is trapped,}\\ \text{it means that that it is in equilibrium under the influence of the above three forces, i.e.,}\\ F_{2} = F_{T} + F_{1}\\ \Rightarrow P_2(\pi R^2) = 2\pi R T + P_1(\pi R^2)\\ \Rightarrow (P_2 -P_1)\pi R^2 = 2\pi RT\\ \therefore R = \dfrac{2T}{P_2-P_1}$

Imagine that you’re on a Triton submersible as a part of a marine biology expedition. Suppose that your room has a circular window hatch with a diameter of $1\,m$ that can be used to access the outside water. If the submersible is internally pressurized at $1\,atm$ and the centre of the hatch is $20\,m$ below sea level then determine the force with which the hatch needs to be pushed in order to open it. 

Determine the area of the hatch that will be subjected to the force first, then measure the magnitude of the force required.

 $\text{Given that}\,r=0.5\,m\,h = 20\,m\,\text{and}\,P=1\,atm = 101325\,Nm^{-2}\\ \text{The area of the hatch will be}\,\pi r^2 = \pi \times 0.5^2 = 0.25\pi\\ \text{Therefore, the force needed to open the hatch will be:}\\ F = P \times A\\ \Rightarrow F = 101325 \times 0.25\pi\\ \therefore F = 7.95 \times 10^{4}\,N \approx 8 \times 10^{4}\,N$

Hydraulic braking systems are used in almost all passenger vehicles and light trucks. They essentially use a brake fluid which is glycol-ether based that runs through the cylinders and pipes of the braking system. When the brake pedal is pressed, a pushrod transfers the force from the pedal to the master cylinder's piston, allowing the brake fluid to be forced into a wheel cylinder near the wheel.The fluid entering the wheel cylinder exerts a force on the pistons, which in turn exerts a force on the brake shoes, which press against the wheel rim to stop the wheel from spinning and bring the vehicle to a halt.From the figure given below, determine the force $F^{\prime}$ that is exerted on the brake shoes.What would this force be if the radius of the wheel cylinder is $\,2.5$ times the radius of the master cylinder and the pushrod actuated 70% of the applied force in the horizontal direction?<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/eb991604-a789-450b-9bb2-6c79ed9fc0b04050664413399325501.png"/>

We know that any force applied to one part of a closed fluid system is transmitted uniformly and without loss to any point of the fluid and the container's walls.

$\text{Considering the force exerted by the pistons of the master}\\ \text{and wheel cylinders, we have, from Pascal’s law:}\\ \dfrac{F_{master}}{A_{master}} = \dfrac{F_{wheel}}{A_{wheel}}\\ \Rightarrow F_{wheel} = \dfrac{A_{wheel}}{A_{master}}.F_{master}\\ \text{Plugging in the appropriate variables,}\\ \text{we find the force exerted by the wheel cylinder pistons on the brake shoes to be:}\\ F^{\prime} = \dfrac{\pi r_2^2}{\pi r_1^2} \times F\\ \Rightarrow F^{\prime} = \dfrac{r_2^2}{r_1^2}F\\ \text{Now, we are given that}\,r_2 = 2.5r_1\,\text{and}\\ F = \dfrac{70}{100} \times F_{applied} = 0.7F_{applied}\\ \text{Therefore, the force on the brake shoes will now be:}\\ F^{\prime} = \dfrac{(2.5r_1)^2}{r_1^2} \times 0.7F_{applied}\\ \therefore F^{\prime} = 4.375F_{applied}$