A vessel containing water is given a constant acceleration a towards the right along a straight hori

Q: Why is the water surface a straight line and not curved like a parabola?

A curved parabolic surface occurs when a vessel rotates because the centrifugal pseudo force depends on the distance from the axis of rotation ($m\omega^2 r$), making the force non-uniform. In contrast, for linear acceleration, the pseudo force ($ma$) is uniform everywhere, resulting in a constant tilt angle and thus a straight line surface.

Q: What happens to the tilt angle if the acceleration of the vessel is equal to $g$?

If $a = g$, then $\tan\theta = a/g = 1$, which means the water surface will tilt at exactly $45^\circ$ to the horizontal.

Q: Why must the liquid surface be perpendicular to the net force?

If the surface were not perpendicular, a component of the net force would act parallel to the surface. Since liquids cannot support shear stress in static equilibrium, this tangential force would cause the liquid to flow. Thus, relative static equilibrium requires the surface to be perpendicular to the net force.

Q: Does the density of the liquid affect the angle of tilt?

No, the density of the liquid does not affect the tilt angle. In the formula $\tan\theta = ma/mg = a/g$, the mass $m$ of the particle cancels out, meaning the tilt depends solely on the horizontal acceleration $a$ and the acceleration due to gravity $g$.

Q: What happens if the vessel is in free fall vertically while accelerating horizontally?

If the vessel is in free fall, the effective vertical acceleration due to gravity becomes zero ($g_{eff} = 0$). In this case, the net force is purely horizontal (due to the pseudo force), and the liquid would try to align itself vertically along the walls, or spill out if the container is open.

Question

A vessel containing water is given a constant acceleration $a$ towards the right along a straight horizontal path. Which of the following diagrams represents the surface of the liquid?

Options:

(A) Tilted upwards to the right

(B) Curved surface

(C) Tilted downwards to the right (straight line)

(D) None of these

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are holding a rectangular glass vessel filled with water.

If you stand still, the surface of the water is perfectly horizontal.

This is because the only force acting on the water particles is gravity, which pulls straight down.

But what happens when you start running to the right with a constant acceleration $a$?

Suddenly, the water sloshes backward, piling up against the left wall of the vessel.

Why does this happen, and what is the exact shape of the water surface during this accelerated motion?

To analyze this, let's step into the **non-inertial frame of reference** of the accelerating vessel.

### The Forces at Play

From the perspective of an observer moving with the vessel, any object of mass $m$ experiences a **pseudo force** in the direction opposite to the acceleration.

Since the vessel accelerates to the right with acceleration $a$, the pseudo force $\vec{F}_{pseudo}$ on a water particle of mass $m$ acts horizontally to the left:

$$\vec{F}_{pseudo} = -m\vec{a}$$

At the same time, the real gravitational force $\vec{F}_g$ acts vertically downwards:

$$\vec{F}_g = m\vec{g}$$

Therefore, the net effective force $\vec{F}_{net}$ acting on any water particle on the surface is the vector sum of these two perpendicular forces:

$$\vec{F}_{net} = \vec{F}_g + \vec{F}_{pseudo}$$

This net force points diagonally downwards and to the left.

### The Free Surface Condition

In fluid statics, a fundamental rule is that **the free surface of a liquid in equilibrium must always be perpendicular to the net effective force** acting on its particles.

If the surface were not perpendicular, there would be a component of the net force acting parallel (tangential) to the surface.

Since fluids cannot withstand shear stress without flowing, this tangential force would cause the water to move.

Because the water is in a state of relative equilibrium inside the vessel, no such flow occurs, meaning the surface must be perfectly perpendicular to $\vec{F}_{net}$.

Let's calculate the angle $	heta$ that the net force vector makes with the vertical:

$$	an	heta = \frac{F_{pseudo}}{F_g} = \frac{ma}{mg} = \frac{a}{g}$$

Since the net force vector is tilted at an angle $	heta$ to the vertical, the perpendicular free surface of the liquid must also be tilted at the exact same angle $	heta$ to the horizontal.

### Concluding the Shape

Because the acceleration $a$ is constant, the angle $	heta$ is constant for every single point on the surface.

This means the free surface of the liquid is a **straight line** tilted at an angle:

$$	heta = 	an^{-1}\left(\frac{a}{g}ight)$$

Since the pseudo force pushes the water to the left, the water level is higher on the left (rear) side and lower on the right (front) side.

This perfectly matches the representation in **Option (c)**.

Animated Solution for Physics - Properties of Solids and Liquids: A vessel containing water is given a constant acceleration $a$ towards the right along a straight horizontal path. Which of the following diagrams represents the surface of the liquid?

Select Answer:

Visualized Solution

Visualizing the Accelerating Vessel

Introducing the Pseudo Force

Accounting for Gravity

Finding the Net Effective Force

Calculating the Angle of Net Force

The Free Surface Condition

Determining the Tilt of the Surface

Selecting the Correct Diagram

The Sigma Insight: Fluid Pressure and Pascal's Law

Analyzing the Setup

The Forces at Play

The Free Surface Condition

Concluding the Shape

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There is a circular tube in a vertical plane. Two liquids which do not mix and of densities $d_{1}$ and $d_{2}$ are filled in the tube. Each liquid subtends $9 0^{\circ}$ angle at centre. Radius joining their interface makes an angle $α$ with vertical. Ratio $\frac{d _{1}}{d _{2}}$ is

Select Answer:

Visualized Solution

Visualizing the Accelerating Vessel

Introducing the Pseudo Force

Accounting for Gravity

Finding the Net Effective Force

Calculating the Angle of Net Force

The Free Surface Condition

Determining the Tilt of the Surface

Selecting the Correct Diagram

The Sigma Insight: Fluid Pressure and Pascal's Law

Analyzing the Setup

The Forces at Play

The Free Surface Condition

Concluding the Shape

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