A large open top container of negligible mass and uniform cross-sectional area A has a small hole of

Q: Why does the container move if there is no external force acting on the system?

While there is no external horizontal force on the entire system (liquid + container), there is an internal separation of momentum. The liquid escaping to the right carries momentum, which exerts an equal and opposite reaction force (thrust) to the left on the container. This is exactly how rockets work!

Q: Why is the acceleration of the container constant in part (a) if the liquid is continuously draining?

At the very instant $t = 0$, the mass of the liquid is $m_0$ and the thrust force is $F = \frac{m_0 g}{50}$. The acceleration at that instant is $a = \frac{F}{m_0} = \frac{g}{50}$. As time progresses, both the mass of the liquid and the thrust force decrease proportionally, keeping the instantaneous acceleration constant as long as the height is uniform!

Q: Can we use Bernoulli's equation to find the velocity of efflux?

Absolutely! Applying Bernoulli's equation between the top surface of the liquid and the outlet of the hole yields $P_0 + \rho g H + 0 = P_0 + \frac{1}{2}\rho v^2$, which simplifies directly to Torricelli's Law: $v = \sqrt{2gH}$.

Q: Why is the mass of the container assumed to be negligible?

Assuming a massless container simplifies the calculation by ensuring that the total mass of the accelerating system is purely the mass of the remaining liquid. If the container had mass $M_c$, the acceleration would be $a = \frac{F}{M_c + m(t)}$, which would vary with time.

Q: What happens to the acceleration if the floor has friction?

If the floor has a coefficient of static friction $\mu_s$, the container will only start moving if the thrust force exceeds the maximum static friction force, i.e., $F_{\text{thrust}} > \mu_s m_0 g$. This requires $\mu_s < 0.02$.

Question

A large open top container of negligible mass and uniform cross-sectional area $A$ has a small hole of cross-sectional area $A/100$ in its side wall near the bottom. The container is kept on a smooth horizontal floor and contains a liquid of density $\rho$ and mass $m_0$. Assuming that the liquid starts flowing out horizontally through the hole at $t = 0$. Calculate
(a) the acceleration of the container and
(b) velocity of efflux when 75% of the liquid has drained out.

Sigma Percentile · Accepted Answer

### Introduction

Imagine a large bucket filled with water resting on a perfectly smooth, frictionless floor.

Suddenly, you punch a tiny hole near the bottom.

Water shoots out horizontally to the right.

What happens to the bucket?

It starts sliding to the left!

This is not magic; it is the fundamental principle of **momentum conservation** and **rocket propulsion** in action.

In this article, we will dive deep into this classic JEE Advanced problem and uncover the elegant physics behind it.

### Analyzing the Setup

Let's break down the physical parameters given to us:

- The container has a uniform cross-sectional area $A$ and its mass is negligible.
- It contains a liquid of density $ho$ and total mass $m_0$.
- A small hole of area $a = \frac{A}{100}$ is made in its side wall near the bottom.
- The floor is perfectly smooth (frictionless).

First, we need to find the initial height $H$ of the liquid column.

Since the container is a cylinder of uniform cross-sectional area $A$, the volume of the liquid is:

$$V = A \cdot H$$

Using the definition of density, the mass of the liquid is:

$$m_0 = V \cdot ho = A \cdot H \cdot ho$$

From this, we can express the initial height $H$ as:

$$H = \frac{m_0}{A ho}$$

### The Magic of Torricelli's Law

When a small hole is opened at a depth $H$ below the free surface of a liquid, the velocity of efflux $v$ is given by **Torricelli's Law**:

$$v = \sqrt{2gH}$$

Substituting our expression for $H$ into this formula, we get the initial velocity of efflux:

$$v = \sqrt{2g \left(\frac{m_0}{A ho}ight)} = \sqrt{\frac{2 g m_0}{A ho}}$$

This is the speed at which the liquid rushes out horizontally at $t = 0$.

### The Birth of Thrust

As the liquid leaves the container, it carries momentum to the right.

According to Newton's Third Law, the escaping liquid must exert an equal and opposite reaction force—known as **thrust**—to the left on the container.

Let's derive the formula for this thrust force $F$ from first principles.

In a small time interval $dt$, the mass of liquid escaping through the hole is:

$$dm = ho \cdot dV = ho \cdot (a \cdot dx) = ho \cdot a \cdot (v \cdot dt)$$

Therefore, the rate of mass flow is:

$$\frac{dm}{dt} = ho \cdot a \cdot v$$

Since this mass leaves with velocity $v$, the rate of change of momentum (which is the thrust force $F$) is:

$$F = v \cdot \frac{dm}{dt} = ho \cdot a \cdot v^2$$

This is a beautiful and universal result in fluid dynamics!

### Part (a): Calculating the Acceleration

Now, let's substitute our parameters into the thrust force equation:

- The area of the hole is $a = \frac{A}{100}$.
- The square of the velocity of efflux is $v^2 = \frac{2 g m_0}{A ho}$.

Plugging these in:

$$F = ho \cdot \left(\frac{A}{100}ight) \cdot \left(\frac{2 g m_0}{A ho}ight)$$

Notice how beautifully the terms cancel out!

The density $ho$ and the area $A$ vanish completely, leaving us with:

$$F = \frac{2 g m_0}{100} = \frac{m_0 g}{50}$$

Using Newton's Second Law, the acceleration of the container at $t = 0$ is:

$$a_{	ext{container}} = \frac{F}{m_0} = \frac{\frac{m_0 g}{50}}{m_0} = \mathbf{\frac{g}{50}}$$

This is a constant acceleration, independent of the density of the liquid or the area of the container!

### Part (b): The Draining Liquid

Next, we want to find the velocity of efflux when **75% of the liquid has drained out**.

If 75% of the liquid has drained, then the remaining mass of the liquid inside the container is:

$$m_{	ext{remaining}} = 25\% 	ext{ of } m_0 = \frac{m_0}{4}$$

Let the new height of the liquid column be $h$.

Since the cross-sectional area $A$ is uniform, the remaining mass is:

$$m_{	ext{remaining}} = A \cdot h \cdot ho = \frac{m_0}{4}$$

Solving for the new height $h$:

$$h = \frac{m_0}{4 A ho} = \frac{H}{4}$$

As expected, the height of the liquid column has dropped to exactly one-fourth of its initial value.

### Final Calculation for Part (b)

Using Torricelli's Law for this new height $h$, the new velocity of efflux $v'$ is:

$$v' = \sqrt{2gh}$$

Substituting $h = \frac{m_0}{4 A ho}$:

$$v' = \sqrt{2g \left(\frac{m_0}{4 A ho}ight)} = \mathbf{\sqrt{\frac{g m_0}{2 A ho}}}$$

This is our final answer for part (b)!

### Socratic Reflection & The Way Forward

Let's take a moment to appreciate the elegance of this system.

Even though the mass of the liquid is continuously decreasing, the instantaneous acceleration of the container remains constant at $g/50$ as long as the liquid level is uniform and the container is massless.

Why?

Because both the remaining mass and the thrust force decrease at the exact same rate!

What if the floor had a coefficient of static friction $\mu_s$?

For the container to start moving at all, the initial thrust force must overcome the maximum static friction:

$$F_{	ext{thrust}} > f_{s,	ext{max}} \implies \frac{m_0 g}{50} > \mu_s m_0 g \implies \mu_s < \frac{1}{50} = 0.02$$

If the friction coefficient is greater than $0.02$, the container will remain stubbornly at rest.

This is how we build true physical intuition—by asking "what if" and exploring the boundaries of our equations!

Visualized Solution

Visualizing the Setup

Connecting Mass and Height

Finding the Initial Height $H$

Velocity of Efflux $v$

The Origin of Thrust Force

Substituting the Parameters

Calculating the Thrust Force $F$

Finding the Acceleration

Analyzing the 75% Drained State

Finding the New Height $h$

Calculating the New Velocity of Efflux $v^{'}$

Summary of Results

The Way Forward

The Sigma Insight: Flow of Fluid

Introduction

Analyzing the Setup

The Magic of Torricelli's Law

The Birth of Thrust

Part (a)

Part (b)

Final Calculation for Part (b)

Socratic Reflection & The Way Forward

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