Water from a tap emerges vertically downwards with an initial speed of 1.0 m/s. The cross-sectional

Q: Why can we assume the pressure is constant throughout the stream?

Because the stream of water is falling freely in the open air, every part of its outer surface is in direct contact with the atmosphere. Therefore, the pressure at any point in the stream is equal to the atmospheric pressure $P_0$.

Q: Can we use $g = 9.8\text{ m/s}^2$ instead of $10\text{ m/s}^2$?

Yes, but using $g = 10\text{ m/s}^2$ gives a perfect square ($4.0$) under the square root, making the calculation extremely clean and matching the options perfectly. In JEE, if the options are widely spaced, using $10\text{ m/s}^2$ is a safe and fast approximation.

Q: What would happen if the fluid was highly viscous?

If the fluid had significant viscosity, there would be internal viscous drag forces that dissipate mechanical energy as heat. In that case, the velocity $v_2$ would be less than $2.0\text{ m/s}$, and consequently, the stream would not narrow down as much.

Q: Why does the stream eventually break into droplets further down?

As the stream narrows, surface tension forces become dominant. Due to Plateau-Rayleigh instability, a long cylinder of fluid minimizes its surface energy by breaking up into spherical droplets. This is why falling water eventually turns into a spray.

Q: Is the equation of continuity valid for gases as well?

The general equation of continuity ($\rho_1 A_1 v_1 = \rho_2 A_2 v_2$) is always valid for any fluid due to conservation of mass. However, because gases are highly compressible, their density $\rho$ changes with pressure and temperature, so $A_1 v_1 = A_2 v_2$ only holds if the flow is subsonic and density changes are negligible.

Question

Water from a tap emerges vertically downwards with an initial speed of $1.0	ext{ m/s}$. The cross-sectional area of tap is $10^{-4}	ext{ m}^2$. Assume that the pressure is constant throughout the stream of water and that the flow is steady, the cross-sectional area of stream $0.15	ext{ m}$ below the tap is

Options:

(A) $5.0 	imes 10^{-4}	ext{ m}^2$

(B) $1.0 	imes 10^{-4}	ext{ m}^2$

(C) $5.0 	imes 10^{-5}	ext{ m}^2$

(D) $2.0 	imes 10^{-5}	ext{ m}^2$

Sigma Percentile · Accepted Answer

### The Magic of Falling Water

Have you ever stood by a kitchen tap, turned it on gently, and watched the stream of water as it falls?

If you look closely, you will notice a beautiful, elegant phenomenon: the stream is wide at the mouth of the tap, but it narrows down into a thin, delicate thread of water as it plunges downwards.

This is not a random occurrence; it is a stunning demonstration of the fundamental laws of fluid mechanics in action!

In this article, we will dive deep into a classic problem from the prestigious JEE archives that quantifies this exact phenomenon.

Let's embark on this thrilling journey to understand how gravity and fluid continuity work hand-in-hand to shape the world around us.

### Analyzing the Setup

Let's first write down the physical parameters given to us in the problem.

We are dealing with a steady, streamline flow of water emerging vertically downwards from a tap.

At the exit of the tap, which we will designate as **Section 1**, we have:
- The initial cross-sectional area: $A_1 = 10^{-4} 	ext{ m}^2$
- The initial velocity of the water: $v_1 = 1.0 	ext{ m/s}$

We are asked to find the cross-sectional area of the stream, $A_2$, at a distance $h = 0.15 	ext{ m}$ below the tap, which we will call **Section 2**.

To solve this, we need to bring in two of the most powerful tools in fluid dynamics: the **Equation of Continuity** and the **Principle of Conservation of Energy**.

### The Master Equation: Continuity of Flow

Why does the water stream narrow down as it falls?

The answer lies in the fact that water is an **incompressible fluid**.

This means its density, $ho$, remains constant throughout the flow.

Since no mass is being created or destroyed, the mass of water entering Section 1 per second must be exactly equal to the mass of water leaving Section 2 per second.

This conservation of mass is mathematically expressed as the **Equation of Continuity**:

$$A_1 v_1 = A_2 v_2$$

From this equation, we can express our target variable, the final area $A_2$, as:

$$A_2 = A_1 \left( \frac{v_1}{v_2} ight)$$

This simple yet profound equation tells us that if the velocity of the water increases as it falls, the cross-sectional area must decrease proportionally to keep the flow rate constant.

Therefore, our immediate goal is to find the final velocity, $v_2$, at Section 2.

### Unleashing Gravity: Conservation of Energy

How do we find the velocity of a fluid element after it has fallen through a height $h$?

Since the stream is open to the atmosphere, the pressure at both Section 1 and Section 2 is equal to the atmospheric pressure, $P_0$.

With no pressure difference to do work on the fluid, the only force acting on the falling water is gravity.

This allows us to treat any small packet of water as a freely falling body!

Applying the third equation of motion (or Bernoulli's equation with constant pressure), we get:

$$v_2^2 = v_1^2 + 2gh$$

This equation is a direct consequence of the **Conservation of Mechanical Energy**, where the loss in gravitational potential energy is converted entirely into kinetic energy.

### Step-by-Step Calculation

Let's substitute the given values into our energy equation to find $v_2$.

Using the standard value for acceleration due to gravity, $g = 10 	ext{ m/s}^2$:

$$v_2^2 = (1.0)^2 + 2 	imes 10 	imes 0.15$$

Let's simplify the terms:

$$v_2^2 = 1.0 + 3.0$$

$$v_2^2 = 4.0 	ext{ m}^2	ext{/s}^2$$

Taking the square root of both sides, we find the velocity of the water at Section 2:

$$v_2 = 2.0 	ext{ m/s}$$

The water has accelerated from $1.0 	ext{ m/s}$ to $2.0 	ext{ m/s}$ over a fall of just $15 	ext{ cm}$!

### Finding the Final Area

Now that we have the final velocity, we can plug it back into our continuity equation to find the new cross-sectional area, $A_2$:

$$A_2 = 10^{-4} 	imes \left( \frac{1.0}{2.0} ight)$$

$$A_2 = 0.5 	imes 10^{-4} 	ext{ m}^2$$

To match the standard scientific notation of the options, we rewrite this as:

$$A_2 = \mathbf{5.0 	imes 10^{-5} 	ext{ m}^2}$$

This is our final result!

Comparing this with the given options, we see that it perfectly matches **Option (c)**.

### The Way Forward: Deepening Your Intuition

What makes this problem so beautiful is how it connects simple kinematics with fluid dynamics.

In more advanced JEE problems, you might encounter situations where the fluid is flowing inside a closed, tapering pipe.

In such cases, the pressure is *not* constant, and you must use the full Bernoulli's equation:

$$P_1 + \frac{1}{2}ho v_1^2 + ho g h_1 = P_2 + \frac{1}{2}ho v_2^2 + ho g h_2$$

Understanding the transition from free-falling streams to pressurized pipe flows is the key to mastering fluid mechanics.

Keep visualizing, keep questioning, and you will find that physics is happening all around you, even in the simple flow of water from a tap!

Select Answer:

Visualized Solution

Visualizing the Flow

Listing the Knowns

The Equation of Continuity

Applying Conservation of Energy

Substituting Values for Velocity

Computing the Velocity $v_{2}$

Substituting into the Continuity Equation

Calculating the Final Area

Matching with the Options

Exploring Further Variations

The Sigma Insight: Flow of Fluid

The Magic of Falling Water

Analyzing the Setup

The Master Equation

Unleashing Gravity

Step-by-Step Calculation

Finding the Final Area

The Way Forward

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