Consider a horizontally oriented syringe containing water located at a height of 1.25 m above the gr

Q: Why did we use the continuity equation instead of Bernoulli's equation to find the exit velocity?

The continuity equation is a direct consequence of mass conservation for an incompressible fluid and must always hold true. Bernoulli's equation relates pressure, elevation, and velocity, but since we are given the exact speed of the plunger and the cross-sectional areas, the continuity equation directly and uniquely determines the exit velocity without needing pressure information.

Q: Does the density of the liquid affect the horizontal range in this problem?

No, the density of the liquid does not affect the range here. The exit velocity is determined purely by the geometry of the syringe and the plunger speed (via continuity), and the time of flight under gravity is independent of mass (and thus density). However, if viscosity or friction were considered, density and viscosity would play a role.

Q: Why is the initial vertical velocity of the water zero?

The syringe is oriented horizontally, meaning the nozzle points parallel to the ground. Therefore, the water is ejected purely in the horizontal direction, making its initial vertical velocity component $u_y = 0$.

Q: What would happen to the range if we halved the nozzle diameter?

If the nozzle diameter is halved, the ratio $D_1/D_2$ doubles. Since the exit velocity depends on the square of this ratio, $v_2$ would increase by a factor of $4$. Since the time of flight remains unchanged, the horizontal range would also increase by a factor of $4$ (becoming $8\text{ m}$).

Q: Is the path of the water stream a perfect parabola?

Yes, in the absence of air resistance, each water droplet acts as a projectile under uniform gravity, which mathematically traces a perfect parabolic path described by the equation $y = -\frac{g}{2 v_2^2} x^2$.

Question

Consider a horizontally oriented syringe containing water located at a height of $1.25	ext{ m}$ above the ground. The diameter of the plunger is $8	ext{ mm}$ and the diameter of the nozzle is $2	ext{ mm}$. The plunger is pushed with a constant speed of $0.25	ext{ m/s}$. Find the horizontal range of water stream on the ground. (Take $g = 10	ext{ m/s}^2$).

Sigma Percentile · Accepted Answer

### Introduction to the Syringe Problem

Imagine holding a simple medical syringe filled with water.

When you push the plunger, water squirts out of the tiny nozzle.

Have you ever wondered how the speed of your hand relates to how far the water travels?

This classic problem from the **JEE Advanced 2004** exam beautifully bridges two fundamental areas of physics: **Fluid Dynamics** (specifically, the Principle of Continuity) and **Kinematics** (Projectile Motion).

Let's dive deep into the mechanics of this system and discover how a simple push translates into a parabolic trajectory.

---

### Analyzing the Setup

We are given a horizontally oriented syringe containing water, positioned at a height of $1.25	ext{ m}$ above the ground.

The syringe has two distinct regions with different diameters:
1. The **plunger (barrel)** with a diameter $D_1 = 8	ext{ mm}$.
2. The **nozzle** with a diameter $D_2 = 2	ext{ mm}$.

The plunger is pushed forward at a constant speed of $v_1 = 0.25	ext{ m/s}$.

Our goal is to find the horizontal distance $R$ (the range) where the water stream hits the ground.

---

### Step 1: Finding the Exit Velocity (The Principle of Continuity)

Before we can calculate how far the water travels, we must determine how fast it leaves the nozzle.

Since water is an incompressible fluid, the volume of water entering any section per second must equal the volume of water leaving it. This is known as the **Principle of Continuity**:

$$A_1 v_1 = A_2 v_2$$

where:
* $A_1$ is the cross-sectional area of the plunger.
* $v_1$ is the velocity of the plunger.
* $A_2$ is the cross-sectional area of the nozzle.
* $v_2$ is the exit velocity of the water.

Since the cross-sections are circular, we can express the areas in terms of their diameters $D_1$ and $D_2$:

$$A_1 = \frac{\pi D_1^2}{4} \quad 	ext{and} \quad A_2 = \frac{\pi D_2^2}{4}$$

Substituting these into the continuity equation:

$$\frac{\pi D_1^2}{4} v_1 = \frac{\pi D_2^2}{4} v_2$$

Notice how the constant factor $\frac{\pi}{4}$ cancels out from both sides, leaving us with a highly elegant relationship:

$$D_1^2 v_1 = D_2^2 v_2$$

Now, let's solve for the exit velocity $v_2$:

$$v_2 = v_1 \left(\frac{D_1}{D_2}ight)^2$$

Substituting the given values ($D_1 = 8	ext{ mm}$, $D_2 = 2	ext{ mm}$, and $v_1 = 0.25	ext{ m/s}$):

$$v_2 = 0.25 	imes \left(\frac{8}{2}ight)^2$$

$$v_2 = 0.25 	imes 16 = 4	ext{ m/s}$$

Thus, the water exits the nozzle horizontally at a speed of **$4	ext{ m/s}$**.

---

### Step 2: Kinematics of the Water Stream (Projectile Motion)

Once the water leaves the nozzle, it is no longer confined by the syringe walls. It is in free fall under gravity.

Since the nozzle is horizontal, the water stream behaves exactly like a horizontally launched projectile.

We can analyze its motion by splitting it into independent vertical and horizontal components.

#### Vertical Motion (Finding the Time of Flight)

In the vertical direction, the water starts with zero initial velocity because it was launched horizontally:

* Initial vertical velocity, $u_y = 0$
* Vertical displacement, $y = -h = -1.25	ext{ m}$ (downwards)
* Vertical acceleration, $a_y = -g = -10	ext{ m/s}^2$

Using the second equation of motion:

$$y = u_y t + \frac{1}{2} a_y t^2$$

Substituting our values:

$$-1.25 = 0 - \frac{1}{2} (10) t^2$$

$$-1.25 = -5 t^2$$

$$t^2 = \frac{1.25}{5} = 0.25$$

Taking the square root:

$$t = 0.5	ext{ s}$$

It takes exactly **$0.5	ext{ seconds}$** for any given droplet of water to fall from the nozzle to the ground.

---

### Step 3: Horizontal Motion (Finding the Range)

Now, let's look at the horizontal direction.

Since we neglect air resistance, there are no horizontal forces acting on the water droplets. Therefore, there is no horizontal acceleration ($a_x = 0$).

The horizontal velocity remains constant throughout the flight:

$$v_x = v_2 = 4	ext{ m/s}$$

The horizontal range $R$ is simply the horizontal distance traveled during the time of flight $t$:

$$R = v_x 	imes t$$

Substituting our values:

$$R = 4	ext{ m/s} 	imes 0.5	ext{ s} = 2	ext{ m}$$

### Conclusion

The horizontal range of the water stream on the ground is exactly **$2	ext{ m}$**.

This problem beautifully demonstrates how a slow, controlled push of a plunger ($0.25	ext{ m/s}$) can be amplified into a high-speed jet ($4	ext{ m/s}$) simply by narrowing the exit nozzle, allowing the fluid to shoot out and cover a significant horizontal distance.

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

The Principle of Continuity

Setting up the Continuity Equation

Calculating Exit Velocity $v_{2}$

Analyzing Vertical Motion

Setting up the Time of Flight Equation

Solving for Time of Flight $t$

Analyzing Horizontal Motion

Calculating the Horizontal Range $R$

The Way Forward

The Sigma Insight: Flow of Fluid

Introduction to the Syringe Problem

Analyzing the Setup

Step 1

Step 2

# Vertical Motion (Finding the Time of Flight)

Step 3

Conclusion

Similar Questions

Water is filled in a cylindrical container to a height of $3 m$ . The ratio of the cross-sectional area of the orifice and the beaker is $0.1$ . The square of the speed of the liquid coming out from the orifice is ( $g = 10 m/s^{2}$ )

The water is filled upto height of $12 m$ in a tank having vertical sidewalls. A hole is made in one of the walls at a depth $h$ below the water level. The value of $h$ for which the emerging stream of water strikes the ground at the maximum range is ...... m.

Water is flowing continuously from a tap having an internal diameter $8 \times 1 0^{- 3} m$ . The water velocity as it leaves the tap is $0.4 ms^{- 1}$ . The diameter of the water stream at a distance $2 \times 1 0^{- 1} m$ below the tap is close to

A cylinder of height $20 m$ is completely filled with water. The velocity of efflux of water (in $ms^{- 1}$ ) through a small hole on the side wall of the cylinder near its bottom, is

Water from a tap emerges vertically downwards with an initial speed of $1.0 m/s$ . The cross-sectional area of tap is $1 0^{- 4} m^{2}$ . Assume that the pressure is constant throughout the stream of water and that the flow is steady, the cross-sectional area of stream $0.15 m$ below the tap is

A liquid of density $ρ$ is coming out of a hose pipe of radius $a$ with horizontal speed $v$ and hits a mesh. 50\% of the liquid passes through the mesh unaffected 25\% losses all of its momentum and, 25\% comes back with the same speed. The resultant pressure on the mesh will be

Water flows in a horizontal tube (see figure). The pressure of water changes by $700 Nm^{- 2}$ between $A$ and $B$ , where the area of cross-section are $40 cm^{2}$ and $20 cm^{2}$ , respectively. Find the rate of flow of water through the tube. (Take, density of water = $1000 kgm^{- 3}$ )

The top of a water tank is open to air and its water level is maintained. It is giving out $0.74 m^{3}$ water per minute through a circular opening of $2 cm$ radius is its wall. The depth of the centre of the opening from the level of water in the tank is close to