A container of large uniform cross-sectional area A resting on a horizontal surface, holds two immis

Q: Why does the cylinder experience two different buoyant forces?

Because the cylinder is floating at the interface, part of its volume is submerged in the upper liquid of density $d$, and another part is submerged in the lower liquid of density $2d$. Each liquid exerts an upward buoyant force equal to the weight of the liquid displaced by the cylinder in that specific layer.

Q: Can we find the pressure at the bottom by integrating hydrostatic pressure directly?

Yes, you can! However, it is much more complex because you have to account for the pressure distribution around the floating cylinder. Using the total weight of the entire system (liquids + cylinder) is a brilliant shortcut that avoids these complex boundary calculations.

Q: Why does the atmospheric pressure cancel out in Bernoulli's equation for efflux?

Both the top surface of the liquid and the emerging jet at the hole are open to the atmosphere. Therefore, both points experience atmospheric pressure $p_0$, which cancels out from both sides of the equation.

Q: What happens if the calculated height for maximum range $h_m$ is greater than $H/2$?

If $h_m$ were greater than $H/2$, it would mean the optimum hole location lies in the upper liquid layer. In that case, we would have to re-derive the efflux velocity using only the upper liquid's density $d$, as the hole would no longer be in the denser liquid.

Q: Why does gravity $g$ cancel out in the expression for the horizontal range?

The horizontal range is the product of horizontal velocity (which is proportional to $\sqrt{g}$) and time of flight (which is proportional to $1/\sqrt{g}$). When multiplied, the $\sqrt{g}$ terms cancel out, making the range purely dependent on the geometry of the setup.

Question

A container of large uniform cross-sectional area $A$ resting on a horizontal surface, holds two immiscible, non-viscous and incompressible liquids of densities $d$ and $2d$, each of height $H/2$ as shown in figure. The lower density liquid is open to the atmosphere having pressure $p_0$.

(a) A homogeneous solid cylinder of length $L$ ($L < H/2$), cross-sectional area $A/5$ is immersed such that it floats with its axis vertical at the liquid-liquid interface with length $L/4$ in the denser liquid. Determine
(i) the density $D$ of the solid,
(ii) the total pressure at the bottom of the container.

(b) The cylinder is now removed and the original arrangement is restored. A tiny hole of area $s$ ($s \ll A$) is punched on the vertical side of the container at a height $h$ ($h < H/2$). Determine:
(i) the initial speed of efflux of the liquid at the hole,
(ii) the horizontal distance $x$ travelled by the liquid initially, and
(iii) the height $h_m$ at which the hole should be punched so that the liquid travels the maximum distance $x_m$ initially. Also calculate $x_m$. (Neglect the air resistance in these calculations)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a large container resting on a flat horizontal surface. Inside, it holds two immiscible, non-viscous, and incompressible liquids.

The lower liquid is denser, with a density of $2d$, while the upper liquid is lighter, with a density of $d$. Each liquid layer occupies exactly half the height of the container, which is $\frac{H}{2}$.

Now, we introduce a solid homogeneous cylinder of length $L$ and cross-sectional area $\frac{A}{5}$ into this two-fluid system. The cylinder floats vertically at the interface of the two liquids.

Our goal is to explore the physics of this floating cylinder, calculate the pressure at the bottom of the container, and then analyze the fluid dynamics when a tiny hole is punched in the container's wall.

### Part (a)(i): Finding the Density of the Solid Cylinder

When a body floats in static equilibrium, the net vertical force acting on it must be zero. This means the downward gravitational force (the weight of the cylinder) is perfectly balanced by the upward buoyant force (upthrust).

Since the cylinder spans across both liquids, it experiences two distinct buoyant forces: one from the upper liquid of density $d$, and another from the lower liquid of density $2d$. Let's write down the force balance equation:

$$W = F_{B1} + F_{B2}$$

Here, the weight of the cylinder $W$ is given by:

$$W = V_{	ext{cyl}} D g = \left(\frac{A}{5}ight) L D g$$

where $D$ is the density of the solid cylinder.

The buoyant force $F_{B1}$ exerted by the upper liquid corresponds to the weight of the upper liquid displaced by the cylinder. Since a length of $\frac{3L}{4}$ is submerged in the upper liquid, we have:

$$F_{B1} = \left(\frac{A}{5}ight) \left(\frac{3L}{4}ight) d g$$

Similarly, the buoyant force $F_{B2}$ exerted by the lower liquid corresponds to the weight of the lower liquid displaced by the cylinder. Since a length of $\frac{L}{4}$ is submerged in the lower liquid, we have:

$$F_{B2} = \left(\frac{A}{5}ight) \left(\frac{L}{4}ight) (2d) g$$

Now, substituting these expressions back into our force balance equation:

$$\left(\frac{A}{5}ight) L D g = \left(\frac{A}{5}ight) \left(\frac{3L}{4}ight) d g + \left(\frac{A}{5}ight) \left(\frac{L}{4}ight) (2d) g$$

Notice how beautifully the common terms cancel out! We can divide both sides by $\left(\frac{A}{5}ight) L g$:

$$D = \frac{3}{4}d + \frac{1}{4}(2d)$$

Simplifying this expression:

$$D = \frac{3}{4}d + \frac{2}{4}d = \frac{5}{4}d$$

Thus, the density of the solid cylinder is exactly **$\frac{5}{4}d$**.

### Part (a)(ii): Total Pressure at the Bottom of the Container

To find the total pressure at the bottom of the container, we can use a brilliant physical shortcut. Instead of calculating the hydrostatic pressure layer by layer, we can consider the vertical equilibrium of the entire system (both liquids plus the floating cylinder).

The total downward force at the bottom of the container must equal the atmospheric force acting on the top surface plus the total weight of the liquids and the cylinder:

$$P \cdot A = p_0 \cdot A + W_{	ext{liquids}} + W_{	ext{cylinder}}$$

Let's calculate the individual weights. The weight of the cylinder is:

$$W_{	ext{cylinder}} = \left(\frac{A}{5}ight) L D g = \left(\frac{A}{5}ight) L \left(\frac{5}{4}dight) g = \frac{ALdg}{4}$$

The total weight of the two liquids is the sum of the weights of the upper and lower layers:

$$W_{	ext{liquids}} = \left(A \cdot \frac{H}{2}ight) d g + \left(A \cdot \frac{H}{2}ight) (2d) g = \frac{AHdg}{2} + AHdg = \frac{3}{2}AHdg$$

Now, substituting these weights back into our pressure equation:

$$P \cdot A = p_0 \cdot A + \frac{3}{2}AHdg + \frac{ALdg}{4}$$

Dividing the entire equation by the cross-sectional area $A$ of the container:

$$P = p_0 + \frac{3}{2}hdg + \frac{Ldg}{4} = p_0 + \frac{dg(6H + L)}{4}$$

Thus, the total pressure at the bottom of the container is **$p_0 + \frac{dg(6H + L)}{4}$**.

### Part (b)(i): Initial Speed of Efflux

Now, the cylinder is removed, and the original liquid levels are restored. A tiny hole of area $s$ ($s \ll A$) is punched in the lower half of the container at a height $h$ from the bottom ($h < \frac{H}{2}$).

To find the initial speed of efflux $v$, we apply Bernoulli's theorem between the open top surface of the liquid (Point 1) and the hole (Point 2). Since $s \ll A$, the velocity of the top surface is negligible ($v_1 \approx 0$).

$$P_1 + \frac{1}{2}ho_1 v_1^2 + ho_1 g y_1 = P_2 + \frac{1}{2}ho_2 v_2^2 + ho_2 g y_2$$

At the top surface (Point 1), the pressure is atmospheric pressure $p_0$. The total gauge pressure at the level of the hole is due to the weight of the upper liquid layer of height $\frac{H}{2}$ and density $d$, plus the lower liquid layer of height $\left(\frac{H}{2} - hight)$ and density $2d$:

$$P_{	ext{gauge}} = d g \left(\frac{H}{2}ight) + 2d g \left(\frac{H}{2} - hight)$$

Applying Bernoulli's equation relative to the level of the hole:

$$p_0 + d g \left(\frac{H}{2}ight) + 2d g \left(\frac{H}{2} - hight) = p_0 + \frac{1}{2} (2d) v^2$$

Subtracting $p_0$ from both sides and dividing by $d$:

$$g \frac{H}{2} + 2g \left(\frac{H}{2} - hight) = v^2$$

$$g \frac{H}{2} + g H - 2gh = v^2$$

$$\frac{3}{2}gH - 2gh = v^2$$

$$v = \sqrt{(3H - 4h)\frac{g}{2}}$$

Thus, the initial speed of efflux is **$\sqrt{(3H - 4h)\frac{g}{2}}$**.

### Part (b)(ii): Horizontal Distance Travelled by the Liquid

Once the liquid jet leaves the hole horizontally, it undergoes projectile motion. The vertical distance it must fall to reach the ground is $h$.

Using the equation of motion for the vertical direction:

$$h = \frac{1}{2} g t^2 \implies t = \sqrt{\frac{2h}{g}}$$

The horizontal distance $x$ travelled by the liquid jet is simply the horizontal velocity multiplied by the time of flight:

$$x = v \cdot t = \sqrt{(3H - 4h)\frac{g}{2}} \cdot \sqrt{\frac{2h}{g}}$$

Notice how the acceleration due to gravity $g$ cancels out beautifully:

$$x = \sqrt{(3H - 4h) \cdot h}$$

Thus, the horizontal distance travelled by the liquid initially is **$\sqrt{h(3H - 4h)}$**.

### Part (b)(iii): Maximizing the Horizontal Range

To find the height $h_m$ at which the hole should be punched to maximize the horizontal distance $x$, we can maximize the function inside the square root:

$$f(h) = h(3H - 4h) = 3Hh - 4h^2$$

Differentiating $f(h)$ with respect to $h$ and setting it to zero:

$$\frac{df}{dh} = 3H - 8h = 0 \implies h_m = \frac{3H}{8}$$

Since $h_m = \frac{3H}{8} < \frac{H}{2}$, this point lies within the lower liquid layer, which is consistent with our setup.

To find the maximum horizontal distance $x_m$, we substitute $h_m$ back into our expression for $x$:

$$x_m = \sqrt{\left(\frac{3H}{8}ight)\left(3H - 4\left(\frac{3H}{8}ight)ight)}$$

$$x_m = \sqrt{\left(\frac{3H}{8}ight)\left(3H - \frac{3H}{2}ight)} = \sqrt{\left(\frac{3H}{8}ight)\left(\frac{3H}{2}ight)}$$

$$x_m = \sqrt{\frac{9H^2}{16}} = \frac{3}{4}H$$

Thus, the maximum horizontal distance is **$\frac{3}{4}H$** at a height of **$\frac{3H}{8}$**.

Visualized Solution

Visualizing the Two-Fluid Interface

Translating Equilibrium to Force Balance

Formulating the Buoyancy Equations

Calculating the Density of the Solid

Understanding Pressure at the Bottom

Summing the Weights of the Components

Deriving the Bottom Pressure

Applying Bernoulli's Theorem for Efflux

Calculating the Speed of Efflux

Determining the Horizontal Range

Finding the Height for Maximum Range

The Sigma Insight: Buoyancy and Archimedes' Principle

Analyzing the Setup

Part (a)(i)

Part (a)(ii)

Part (b)(i)

Part (b)(ii)

Part (b)(iii)

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Visualized Solution

Visualizing the Two-Fluid Interface

Translating Equilibrium to Force Balance

Formulating the Buoyancy Equations

Calculating the Density of the Solid

Understanding Pressure at the Bottom

Summing the Weights of the Components

Deriving the Bottom Pressure

Applying Bernoulli's Theorem for Efflux

Calculating the Speed of Efflux

Determining the Horizontal Range

Finding the Height for Maximum Range

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Analyzing the Setup

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