Animated Solution for Physics - Properties of Solids and Liquids: A cylindrical tube, with its base as shown in the figure, is filled with water. It is moving down with a constant acceleration a along a fixed inclined plane with angle θ=45∘. P1 and P2 are pressures at points 1 and 2, respectively, located at the base of the tube. Let β=(P1−P2)/(ρgd), where ρ is density of water, d is the inner diameter of the tube and g is the acceleration due to gravity. Which of the following statement(s) is(are) correct ?
Select Answer:
* Multiple Correct
Visualized Solution
θ=45∘
Tube accelerating down an incline at θ=45∘.
ax,ay
Acceleration components:
ax=acos45∘=2a
ay=asin45∘=2a (downwards)
Δx=d,Δy=d
Introduce Point 3 such that:
Horizontal distance 2→3 is d.
Vertical distance 3→1 is d.
∂x∂P=−ρax
Horizontal pressure gradient:
∂x∂P=−ρax=−ρ2a
P3−P2=−ρ2ad⟹P2−P3=ρ2ad
∂y∂P=−ρ(g−ay)
Vertical pressure gradient:
∂y∂P=−ρ(g−ay)=−ρ(g−2a)
P1−P3=ρ(g−2a)d
P1−P2
(P1−P3)−(P2−P3)=P1−P2
P1−P2=ρ(g−2a)d−ρ2ad
P1−P2=ρd(g−22a)=ρd(g−a2)
β=ρgdP1−P2
β=ρgdP1−P2
β=ρgdρd(g−a2)=1−ga2
β=0,β=22−1
Check options:
If a=2g⟹β=1−gg/2⋅2=0
If a=2g⟹β=1−gg/2⋅2=22−1
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The Sigma Insight: Fluid Pressure and Pascal's Law
Solution Diagram
Analyzing the Setup
Welcome to a beautiful problem from fluid mechanics! Imagine you are standing inside this cylindrical tube, filled with water. But you are not just standing still; you are sliding down a 45∘ incline.
The water inside isn't just experiencing the familiar downward pull of gravity. It is also feeling the effects of this acceleration.
This is the core of the problem: fluid statics in an accelerating reference frame.
When a fluid accelerates, it behaves as if there is an additional "gravity" pulling it in the opposite direction. This is the pseudo force.
The Master Equation
To find the pressure difference between any two points, we need to understand the pressure gradients.
Because the fluid is accelerating down the incline, the acceleration a has two components.
The horizontal component is ax=acos45∘=2a.
The vertical component is ay=asin45∘=2a downwards.
These components create pressure gradients in both the horizontal and vertical directions.
Let's introduce a clever intermediate point, point 3. We place it horizontally aligned with point 2, and vertically aligned with point 1.
Why do we do this? Because it allows us to calculate the horizontal and vertical pressure differences separately!
Horizontal and Vertical Pressure Differences
First, let's look at the horizontal pressure difference between point 2 and point 3.
The fluid is accelerating to the right with ax. This means the pressure must decrease as we move to the right, to provide the net force for this acceleration.
The horizontal pressure gradient is ∂x∂P=−ρax=−ρ2a.
Since the distance between point 2 and point 3 is d, the pressure difference is P2−P3=ρ2ad.
Next, let's tackle the vertical pressure difference between point 3 and point 1.
Here, we must consider both gravity and the vertical component of acceleration.
The tube is accelerating downwards, so the fluid feels an upward pseudo force. This effectively reduces the pull of gravity!
The effective downward acceleration is g−ay=g−2a.
The vertical pressure gradient is ∂y∂P=−ρ(g−2a).
Since point 1 is at a depth d below point 3, the pressure difference is P1−P3=ρ(g−2a)d.
Final Calculation
Now, we have the pieces of the puzzle. We simply subtract the two equations to find the pressure difference between point 1 and point 2.
Notice how the pressure at point 3 beautifully cancels out!
P1−P2=(P1−P3)−(P2−P3)
P1−P2=ρ(g−2a)d−ρ2ad
Simplifying this, we get P1−P2=ρd(g−22a)=ρd(g−a2).
The question asks for the ratio β, which is this pressure difference divided by ρgd.
β=ρgdP1−P2=ρgdρd(g−a2)=1−ga2.
Finally, let's test the given options.
If the acceleration a=2g, then β=1−gg/2⋅2=1−1=0.
This perfectly matches option (A)!
If the acceleration a=2g, then β=1−gg/2⋅2=1−21=22−1.
This perfectly matches option (C)!
And there we have it. By breaking the problem down into horizontal and vertical components, we turned a complex accelerating fluid problem into a simple geometric puzzle.
This is the power of choosing the right reference frame and the right intermediate points!
