Solution for Question 10
Analysis of the Physical Situation:
When a fluid container accelerates, the fluid experiences a pseudo-force in the frame of the container. This alters the effective gravity and the direction of the buoyant force. An air bubble in a liquid always moves opposite to the direction of the effective gravity (or towards the region of lowest pressure).
Step 1: Determining the Direction
Let the acceleration of the tube be $\vec{a}$. In the non-inertial frame of the tube, a pseudo-force acts on every fluid element in the direction $-\vec{a}$.
- The bubble is observed to shift to the right (relative to the highest point).
- The bubble aligns itself opposite to the net effective gravity $\vec{g}_{eff} = \vec{g} – \vec{a}$.
- For the bubble to move right, the fluid must “pile up” on the left, creating higher pressure on the left. This implies the pseudo-force is to the left.
- Therefore, the acceleration $\vec{a}$ of the container must be to the right.
Step 2: Determining the Magnitude
At equilibrium position $\theta$, the tangent to the surface (or the position of the bubble) is perpendicular to the effective gravity vector.
From the vector diagram of forces (Gravity $mg$ acting down, Pseudo-force $ma$ acting left):
$$ \tan \theta = \frac{\text{Horizontal Component}}{\text{Vertical Component}} = \frac{ma}{mg} = \frac{a}{g} $$Solving for the acceleration magnitude $a$:
$$ a = g \tan \theta $$