Tpr q

[AugustMCAT]

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Does a rigid lid not signify that atmospheric pressure should be disregarded? Here's the Q (C is correct answer):


The stopper at the bottom of a tank filled with water has an area of 0.5 m2. The tank has a rigid lid. If the stopper is at a depth of 10 m below the surface, what is the total downward force exerted by the water on the stopper? (Note: The density of water is 1000 kg/m3.)

A.
25,000 N

B.
50,000 N

C.
100,000 N

D.
150,000 N

 

[PingPongPro]

Yeah its in a tank so the atmospheric molecules are not acting on the stopper. If the top was open then you would include atmospheric pressure.

The questions is just= density x gravity x height.

1x10^3 x 10^1 x 10^1= 10^5 = C

 

[dorjiako]

Does a rigid lid not signify that atmospheric pressure should be disregarded? Here's the Q (C is correct answer):


The stopper at the bottom of a tank filled with water has an area of 0.5 m2. The tank has a rigid lid. If the stopper is at a depth of 10 m below the surface, what is the total downward force exerted by the water on the stopper? (Note: The density of water is 1000 kg/m3.)

A.
25,000 N

B.
50,000 N

C.
100,000 N

D.
150,000 N

On first glance, I thought that the atmospheric pressure should be excluded because 1. the tank is filled to the brim, 2. the rigid lid is covering the filled tank of water. However, if the answer is c, then they definitely included the atmospheric pressure in solving this problem. So, I am puzzled as well.

 

[AugustMCAT]

Yeah its in a tank so the atmospheric molecules are not acting on the stopper. If the top was open then you would include atmospheric pressure.

The questions is just= density x gravity x height.

1x10^3 x 10^1 x 10^1= 10^5 = C

This is incorrect. The area is 0.5 m2. They're asking for force, so it's P/A=F. Hence, 50,000N, which is what I arrived at. But in their calculations, TPR included atmospheric pressure, which I found odd, since they explicitly said there was a rigid lid.

 

[PingPongPro]

This is incorrect. The area is 0.5 m2. They're asking for force, so it's P/A=F. Hence, 50,000N, which is what I arrived at. But in their calculations, TPR included atmospheric pressure, which I found odd, since they explicitly said there was a rigid lid.

Oh oops, you are right. I didn't really read the question fully 🙄.

I think its highly likely there may just be a misprint or an error on TPR's part.

 

[AugustMCAT]

Ok cool...so is the general consensus that if there is a rigid convering to a container of water, the water is excluded from atmospheric pressure (assuming the water level is not in contact with the lid)?

 

[Kimchii]

Though to put the rigid lid on requires first for the container to be exposed to air pressure. Unless it were in a vacuum there would still be air pressure. What if you used this logic to solve the problem? Would you arrive at C?

 

[AugustMCAT]

Though to put the rigid lid on requires first for the container to be exposed to air pressure. Unless it were in a vacuum there would still be air pressure. What if you used this logic to solve the problem? Would you arrive at C?

Yes, but this is not the case. The amount of air trapped in the container is negligible (remember, pressure is a function of depth).