Class 11 Physics | Fluid Dynamics | #12 Solved Example-5 on Fluid Dynamics | For JEE & NEET

[@maxwelllearning4923]

Sir ,after calculation answer comes out to be 5 m/s

[@shivamummoju4820]

At 2:27 shouldnt we write the power of radius as 4? When we are substituting the value of Va in the expression the power of radius should be 4?

[@prakhar1999]

yes!!

[@aanchalsharma1298]

Pls tell sir

[@rupaliagarwal603]

Sir, At 2:27 shouldn't we write the power of radius as 4? When we are substituting the value of Va in the expression the power of radius should be 4?

[@ayushsaxena1164]

Dear Sir pls tell whether pressure energy at pt. B should be P-atm + rho*gh or if I am mistaken here. Sir also why we have not considered rho*gh for pt. B here as it is not at ground level so I think it should be mentioned.🤔🤔

[@vishwajeetchoudhary8264]

So, as the answer doesn't depend on density of liquid, so speed will be same for any liquid used ?

[@DAI-_]

Yes maybe...

[@RoshanPradhan2]

I think the answer should be 5.04 m/s considering 0.8^4

[@vibhutimakode5535]

Sir at B pressure should be Patm +pgh

[@khemar-jl1tb]

exactly

[@rashmiverma9677]

Water is not static here its flowing they are not applying any force (Pressure) among themselves

[@harshsameer9440]

bottom surface removed => liquid at B free in free fall and no pressure by liquid layers above B. Therefore P at B = Patm only.

[@ayush8480]

@@harshsameer9440 tysm

[@gourav5423]

Sir why only Patm. I mean as the cross-sectional area is decreasing there should be some difference in the INTERNAL LIQUID PRESSURE ENERGY also?

[@arghadeepbosu1194]

Yes at point B it should be P(atm)+rho×gh

[@urjaagarwal332]

@@arghadeepbosu1194 I think you are wrong. The excess pressure can be viewed as direct effect of mass of the fluid above a particular layer of liquid. If you see, in this example the fluid can be imagined as small sphere like molecules which are all falling down in free fall like motion hence there is no apparent effect of mass of fluid layers above a particular layer. In a closed container the layers pile up on one another like blocks piling on each other and hence the normal force on each of them by the succeeding blocks increases similar to that in case of fluids that can be taken as blocks with negligible height.

[@universalcosmologist3675]

Why everybody is having doubt about Pb Point b is just below base and as base is removed there will be no pressure due to liquid falling Also we have taken reference at B so rhogh for B must be zero right

[@loppol4995]

Sir shouldn't the reassure at point b be patm+rho*g*h

[@physicsgalaxyworld]

+Lop Pol after the bottom is removed its Patm only... before the bottom is removed what you have written is valid...

[@loppol4995]

Thank you sir.

[@AnilVerma-ev1fj]

Dear sir Bernoulli’s equation for A should not have gravitational potential energy(rogh*g*h) as its on top and all the weight is being applied at point B as it is lowest so I guess it should have this term rogh*g*h Plz clear my doubt!!!

[@harshsameer9440]

@@AnilVerma-ev1fj bottom surface removed => liquid at B free in free fall and no pressure by liquid layers above B. Therefore P at B = Patm only.

[@turtlepedia5149]

In early egs we took different pressures

[@GauravKumar-xn8vb]

Sir pressure at point B should be p atm+rho g h ???

[@neelkamal1269]

Because we considered B as a reference point

[@5anjalysanjay197]

Sir, I have the same query, " Why we have considered the Gravitational Potential Energy Term Rho*g*H at A and not at B?

[@adarshchaturvedi3498]

because B is taken as reference zero potential

[@khemar-jl1tb]

but the pressure at b by the liquid above it??

[@deuteriumtritium9700]

@@khemar-jl1tb bottom surface removed => liquid at B free in free fall and no pressure by liquid layers above B. Therefore P at B = Patm only.

[@seemachatterjee4389]

@@deuteriumtritium9700 Thank you for clearing the doubt, I also had the same doubt as arun.

[@ArjunGladiator]

SirWhy we have considered the Gravitational Potential Energy Term Rho*g*H at A and not at B?

[@itzrajaji7064]

Coz h=0 at bottom dear

[@ritviksharma8663]

in this example why pb is equals to patm

[@turtlepedia5149]

Why pressure is same???

[@deuteriumtritium9700]

bottom surface removed => liquid at B free in free fall and no pressure by liquid layers above B. Therefore P at B = Patm only.

[@sarthaksinghchauhan4235]

Why we didn’t take pressure at B =P atm + rho g h

[@deuteriumtritium9700]

bottom surface removed => liquid at B free in free fall and no pressure by liquid layers above B. Therefore P at B = Patm only.

[@prakhar1999]

At 2:27 shouldnt we write the power of radius as 4? When we are substituting the value of Va in the expression the power of radius should be 4?

[@physicsgalaxyworld]

Yes you are right... thanks for pointing it out... will correct and re-upload...

[@faheemking6029]

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[@shivamverma9447]

@@faheemking6029 🤟 nice bro....

[@abhay1576]

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[@abhay1576]

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