Many would have seen this problem… I couldn’t understand the solution and the logic behind the happening of the scenario in the problem … any hints here
When the tube is pulled out , some of the mercury still remains in the tube, like a cork. The atmospheric pressure balances the weight of the mercury and the pressure of the gas inside. Try and find out the pressure of the gas inside the tube(by using P1V1=P2V2 because temperature is constant).
I am getting option A, please confirm.
Got it Thanks!
I can’t figure out how buoyancy will play it’s role here.
Please explain with solution
When the tube is accelerating the direction of g effective will change. Buoyant force acts opposite to direction if g effective. So the bubble aligns itself along that line.
Bro bubble has water on both sides. It’s confusing me how to write the expressions.
When ever you are free please post the solution. ( No peer pressure bro )
You don’t have to worry about the curved surface of the bubble. Buoyant force always acts along g effective and the magnitude also depends only on volume of water displaced.
Since the tube is vertical pressure increases as you go down. And since both sides are open pressure has to be same on both sides. Also there is an excess pressure due to a curved surface ( more pressure towards concave side). Keeping these things mind:
A: Start from top. Pressure just inside is more than Patm. As we go down pressure increases. So the pressure at the lower end is more than Patm. Also the curved surface at the lower end is towards more pressure side. So it’s fine. ( Note that the radius of curvature of both ends won’t be same. It will be smaller at the lower end as the pressure difference is high)
B: Same as A except that the lower end has an opposite curvature. We know this is impossible as pressure inside liquid is more according to the previous argument. But the curvature suggests that pressure outside is more.
C: Now you can tell even C is wrong. Flat surface doesn’t account for any pressure difference.
You can also say that this liquid column is not in equilibrium as it is experiencing buoyant force due to air( in previous cases surface tension was another vertical force apart from buoyancy. So an argument like this will fail there)
D: On your own since @KrishSharma has got it
Ah, I got you know. I was almost going to do an experiment with some glass table in my home lol.
This is like doppler effect I think. If you think like that you will get your answer.
ACD ? Is it ?
Yes I have blurred it.
And I’m here again.
Dk the answer.
The liquid surface will align itself perpendicular to g effective.
Do you have a doubt how to find the direction of g effective? Then I will tell.
Nope. I got it now. Thanks!
Wait how is it not A)? Like let horizontal acceleration be rightwards, then we can take pseudo force to stop the object and now the water has net horizontal acceleration in right direction, which is what we want?
I took components in x and y and solved …take buoyancy in x and y …
X due to acceleration part
Y due to Gravity part and then balanced the normal forces and gravity with it