Water is filled in a cylindrical container to a height of 3 m. The ratio of the cross-sectional area

Q: Why did we not use Torricelli's Law $v = \sqrt{2gh}$ directly?

Torricelli's Law assumes that the container is infinitely wide ($A_1 \gg A_2$), meaning the top surface is practically stationary ($v_1 \approx 0$). Here, the area ratio is given as $0.1$, which is significant, so we must use the exact continuity equation.

Q: How did we find the height $h = 2.475\text{ m}$?

The total height of the water is $3\text{ m}$, and the orifice is at a height of $52.5\text{ cm} = 0.525\text{ m}$ from the bottom. The height of the water column above the orifice is the difference: $3 - 0.525 = 2.475\text{ m}$.

Q: Why are the pressures at the top surface and the orifice both equal to $p_0$?

Both the top surface of the water and the water jet emerging from the orifice are in direct contact with the surrounding air, so they are both at atmospheric pressure ($p_0$).

Q: Does the density of the liquid affect the efflux velocity?

No, as seen in the derivation, the density term $\rho$ cancels out completely from both sides of Bernoulli's equation, meaning the efflux velocity is independent of the liquid's density.

Q: What would happen to the efflux velocity if the beaker was made twice as wide?

If the beaker was wider, the ratio $A_2/A_1$ would decrease, making $v_1$ smaller. The efflux velocity $v_2^2$ would decrease slightly, moving closer to the Torricelli limit of $2gh = 49.5\text{ m}^2/\text{s}^2$.

Question

Water is filled in a cylindrical container to a height of $3	ext{ m}$. 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 ($g = 10	ext{ m/s}^2$)

Options:

(A) $50	ext{ m}^2/	ext{s}^2$

(B) $50.5	ext{ m}^2/	ext{s}^2$

(C) $51	ext{ m}^2/	ext{s}^2$

(D) $52	ext{ m}^2/	ext{s}^2$

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two Levels

Imagine standing next to a massive cylindrical container filled with water to a height of $3	ext{ m}$. Near the bottom, a tiny circular opening—an **orifice**—is punched into the side wall at a height of $52.5	ext{ cm}$ ($0.525	ext{ m}$) from the base.

As water shoots out horizontally from this orifice, the top surface of the water must slowly descend to conserve mass. This is a classic fluid dynamics problem where we cannot simply assume the container is infinitely wide. We must account for the motion of both the escaping jet and the descending top surface.

Our goal is to find the exact value of the **square of the efflux velocity** ($v_2^2$) of the water leaving the orifice.

### The Continuity Connection

Before we look at energy conservation, we must establish how the speed of the descending top surface ($v_1$) relates to the speed of the escaping jet ($v_2$). This relationship is governed by the **Equation of Continuity** for an incompressible fluid:

$$A_1 v_1 = A_2 v_2$$

Here, $A_1$ is the cross-sectional area of the beaker, and $A_2$ is the cross-sectional area of the orifice. Rearranging this equation gives:

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

We are given that the ratio of the cross-sectional area of the orifice to that of the beaker is $\frac{A_2}{A_1} = 0.1$. Substituting this value, we get:

$$v_1 = 0.1 v_2$$

This simple relation tells us that the top surface descends at exactly one-tenth of the speed at which the water shoots out of the orifice.

### Unleashing Bernoulli's Power

To find the velocities, we apply **Bernoulli's Theorem** along a streamline connecting the top surface (Point 1) and the orifice (Point 2):

$$p_1 + ho g y_1 + \frac{1}{2} ho v_1^2 = p_2 + ho g y_2 + \frac{1}{2} ho v_2^2$$

Let's simplify this equation step-by-step:

1. **Pressure:** Both Point 1 and Point 2 are open to the atmosphere, so $p_1 = p_2 = p_0$ (atmospheric pressure). The pressure terms cancel out from both sides.
2. **Reference Level:** Let the bottom of the beaker be our reference level ($y = 0$). Thus, the height of the top surface is $y_1 = H = 3	ext{ m}$, and the height of the orifice is $y_2 = h_0 = 0.525	ext{ m}$.

Substituting these values into Bernoulli's equation yields:

$$ho g H + \frac{1}{2} ho v_1^2 = ho g h_0 + \frac{1}{2} ho v_2^2$$

We can divide the entire equation by the density $ho$:

$$g H + \frac{1}{2} v_1^2 = g h_0 + \frac{1}{2} v_2^2$$

Rearranging the terms to group the gravitational potential energy terms together:

$$g(H - h_0) + \frac{1}{2} v_1^2 = \frac{1}{2} v_2^2$$

Here, $H - h_0$ is the height of the water column directly above the orifice, which we will call $h$:

$$h = 3	ext{ m} - 0.525	ext{ m} = 2.475	ext{ m}$$

So, our simplified equation becomes:

$$g h + \frac{1}{2} v_1^2 = \frac{1}{2} v_2^2$$

### The Final Calculation

Now, we substitute $v_1 = 0.1 v_2$ into our simplified Bernoulli's equation:

$$g h + \frac{1}{2} (0.1 v_2)^2 = \frac{1}{2} v_2^2$$

$$g h + \frac{1}{2} (0.01) v_2^2 = \frac{1}{2} v_2^2$$

Subtracting $\frac{1}{2} (0.01) v_2^2$ from both sides:

$$g h = \frac{1}{2} v_2^2 (1 - 0.01)$$

$$g h = \frac{1}{2} v_2^2 (0.99)$$

Solving for $v_2^2$:

$$v_2^2 = \frac{2 g h}{0.99}$$

Substituting the given values $g = 10	ext{ m/s}^2$ and $h = 2.475	ext{ m}$:

$$v_2^2 = \frac{2 	imes 10 	imes 2.475}{0.99}$$

$$v_2^2 = \frac{49.5}{0.99}$$

$$v_2^2 = 50	ext{ m}^2/	ext{s}^2$$

Thus, the square of the speed of the liquid coming out from the orifice is exactly **$50	ext{ m}^2/	ext{s}^2$**, which corresponds to **Option (a)**.

### Why the Finite Width Matters (The Torricelli Contrast)

If we had used **Torricelli's Law** directly (which assumes an infinitely wide container where $v_1 \approx 0$):

$$v_2^2 \approx 2 g h = 2 	imes 10 	imes 2.475 = 49.5	ext{ m}^2/	ext{s}^2$$

Notice that our exact solution ($50	ext{ m}^2/	ext{s}^2$) is slightly larger! This is because in a container of finite width, the kinetic energy of the descending top surface contributes to the total energy of the system, resulting in a slightly higher efflux velocity to maintain the conservation of energy. This subtle detail is what makes JEE Advanced questions so beautifully precise!

Animated Solution for Physics - Properties of Solids and Liquids: 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 ( $g = 10 m/s^{2}$ )

Select Answer:

Visualized Solution

Understanding the Physical Setup

The Principle of Continuity

Expressing $v_{1}$ in terms of $v_{2}$

Applying Bernoulli's Theorem

Simplifying Bernoulli's Equation

Calculating the Effective Height $h$

Substituting $v_{1}$

Solving for $v_{2}^{2}$

Calculating the Final Value

Torricelli's Approximation vs. Exact Solution

The Sigma Insight: Flow of Fluid

The Setup

The Continuity Connection

Unleashing Bernoulli's Power

The Final Calculation

Why the Finite Width Matters (The Torricelli Contrast)

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