A large open tank has two holes in the wall. One is a square hole of side L at a depth y from the to

Q: Why does the density of the liquid not affect the velocity of efflux?

According to Bernoulli's equation, both the pressure energy and kinetic energy terms are proportional to density, which cancels out when equating them. Thus, the velocity of efflux depends only on the depth of the orifice and gravity.

Q: What if the tank was closed and pressurized at the top?

If the tank were pressurized, the pressure at the top would add an extra term to the velocity of efflux, and the simple Torricelli's formula $\sqrt{2gh}$ would need to be modified to include the pressure difference.

Q: Does the shape of the orifice affect the velocity of efflux?

No, the velocity of efflux is independent of the shape of the orifice. However, the shape determines the cross-sectional area, which directly affects the volume flow rate.

Q: What is the 'Vena Contracta' effect, and why did we neglect it here?

In real fluids, the jet of liquid contracts slightly after passing through an orifice, making the effective area of the jet smaller than the actual area of the hole. In JEE problems, unless specified, we assume ideal fluid flow and neglect this contraction.

Q: Why did the gravity term $g$ cancel out? Does this mean this setup works in zero gravity?

The gravity term cancels because both holes experience the same gravitational field. However, in true zero gravity ($g=0$), there would be no hydrostatic pressure to push the water out, so no efflux would occur unless the tank was pressurized.

Question

A large open tank has two holes in the wall. One is a square hole of side $L$ at a depth $y$ from the top and the other is a circular hole of radius $R$ at a depth $4y$ from the top. When the tank is completely filled with water, the quantities of water flowing out per second from both holes are the same. Then, $R$ is equal to

Options:

(A) $\frac{L}{\sqrt{2\pi}}$

(B) $2\pi L$

(C) $L$

(D) $\frac{L}{2\pi}$

Sigma Percentile · Accepted Answer

### Introduction to Fluid Efflux

Imagine standing next to a massive water reservoir, watching water shoot out from a tiny puncture in its side.

Have you ever wondered what dictates the speed of that emerging stream?

Is it the weight of the water above it?

Is it the shape of the hole?

Or perhaps the density of the liquid itself?

This classic problem from the JEE Advanced 2000 paper invites us to explore these very questions through a beautifully balanced setup involving two orifices of different shapes—one square and one circular—placed at different depths in a water tank.

---

### The Physics of Efflux: Torricelli's Law

To begin our journey, we must first understand how fast water leaves an opening.

In the 17th century, the Italian physicist **Evangelista Torricelli** discovered a remarkable relationship.

He found that the speed of efflux $v$ of an ideal fluid from a small hole at a depth $h$ below the free surface is given by:

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

This is known as **Torricelli's Law**.

Interestingly, this is the exact same speed that a solid object would acquire if it fell freely under gravity from a height $h$.

This elegant result can be derived directly from **Bernoulli's Principle**, which is a statement of the conservation of energy for flowing fluids.

For our setup, we have two different depths:

1. For the square hole at depth $y$, the velocity of efflux is:
   $$v_1 = \sqrt{2gy}$$

2. For the circular hole at depth $4y$, the velocity of efflux is:
   $$v_2 = \sqrt{2g(4y)} = 2\sqrt{2gy}$$

Notice that because the circular hole is four times deeper, the water shoots out of it exactly **twice as fast** as it does from the square hole.

---

### The Equation of Continuity and Volume Flow Rate

Now, how do we quantify the total amount of water leaving each hole per second?

This is where the **Equation of Continuity** comes to our rescue.

The volume of liquid flowing out per second—also known as the **volume flow rate** or **discharge** $Q$—is the product of the cross-sectional area of the opening $A$ and the velocity of efflux $v$:

$$Q = A \cdot v$$

Let's calculate this discharge for both of our holes:

- **The Square Hole:**
  The hole is a square of side $L$, so its area is:
  $$A_1 = L^2$$
  Therefore, its volume flow rate is:
  $$Q_1 = A_1 v_1 = L^2 \sqrt{2gy}$$

- **The Circular Hole:**
  The hole is a circle of radius $R$, so its area is:
  $$A_2 = \pi R^2$$
  Therefore, its volume flow rate is:
  $$Q_2 = A_2 v_2 = \pi R^2 \sqrt{2g(4y)} = 2\pi R^2 \sqrt{2gy}$$

---

### The Elegant Simplification

The problem states a beautiful constraint: **the quantities of water flowing out per second from both holes are the same**.

This means we can set their discharges equal to each other:

$$Q_1 = Q_2$$

Substituting our expressions into this equation:

$$L^2 \sqrt{2gy} = \pi R^2 \sqrt{2g(4y)}$$

Let's simplify the right-hand side by pulling out the constant factor from the square root:

$$L^2 \sqrt{2gy} = \pi R^2 \cdot \sqrt{4} \cdot \sqrt{2gy}$$

$$L^2 \sqrt{2gy} = 2\pi R^2 \sqrt{2gy}$$

Now, look at how beautifully the math simplifies!

The term $\sqrt{2gy}$ is present on both sides and can be completely canceled out:

$$L^2 = 2\pi R^2$$

This leaves us with a wonderfully clean geometric relationship between the side of the square $L$ and the radius of the circle $R$.

---

### The Final Triumph

To find $R$ in terms of $L$, we simply rearrange the equation:

$$R^2 = \frac{L^2}{2\pi}$$

Taking the square root of both sides, we get:

$$R = \frac{L}{\sqrt{2\pi}}$$

This matches **Option (a)** perfectly!

---

### Deepening the Concept: What if?

Let's push our understanding further by asking some Socratic questions:

- **What if the entire tank was placed in an elevator accelerating upwards with acceleration $a$?**
  In this case, the effective acceleration due to gravity becomes $g_{eff} = g + a$. However, since $g$ canceled out completely in our simplification step, the ratio $R = \frac{L}{\sqrt{2\pi}}$ remains **absolutely unchanged**!

- **What if we filled the tank with a denser liquid, like mercury?**
  Since Torricelli's velocity of efflux is independent of the liquid's density, the flow rates would still scale in the exact same way, and the required radius $R$ would remain the same.

Select Answer:

Visualized Solution

Understanding the Setup

Torricelli's Law of Efflux

Volume Flow Rate (Discharge)

Flow Rate for the Square Hole

Flow Rate for the Circular Hole

Equating the Flow Rates

Simplifying the Equation

Solving for $R$

Deepening the Concept

The Sigma Insight: Flow of Fluid

Introduction to Fluid Efflux

The Physics of Efflux

The Equation of Continuity and Volume Flow Rate

The Elegant Simplification

The Final Triumph

Deepening the Concept

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Select Answer:

Visualized Solution

Understanding the Setup

Torricelli's Law of Efflux

Volume Flow Rate (Discharge)

Flow Rate for the Square Hole

Flow Rate for the Circular Hole

Equating the Flow Rates

Simplifying the Equation

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R

Deepening the Concept

The Sigma Insight: Flow of Fluid

Introduction to Fluid Efflux

The Physics of Efflux

The Equation of Continuity and Volume Flow Rate

The Elegant Simplification

The Final Triumph

Deepening the Concept

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