An open-ended U-tube of uniform cross-sectional area contains water (density 10^3 kg m^-3). Initiall

Q: Why do we equate the pressure at the bottom of the U-tube?

According to Pascal's Law, in a continuous, stationary fluid, the pressure at any two points at the same horizontal level must be equal. The bottom of the U-tube is the easiest reference level to use here.

Q: Why is the height of water in the left arm $h_1 - 0.1$?

The problem states that $h_1$ is the total height of the liquid in the left arm. Since the top $0.1 \mathrm{m}$ is kerosene, the remaining height must be water.

Q: What happens to the atmospheric pressure?

Since both arms of the U-tube are open to the atmosphere, atmospheric pressure $P_0$ acts on both sides. When we equate the pressures, $P_0$ cancels out.

Q: Why did we use $0.29 \times 2$ for the initial water height?

Initially, the water level in both arms was $0.29 \mathrm{m}$ from the bottom. So, the total length of the vertical water columns was $0.29 + 0.29 = 0.58 \mathrm{m}$.

Q: Would the answer change if the U-tube had different cross-sectional areas?

Yes! If the areas were different, a drop of $\Delta h$ in the left arm would not result in a rise of $\Delta h$ in the right arm. The volume conservation equation would become $A_1 \Delta h_1 = A_2 \Delta h_2$.

Question

An open-ended U-tube of uniform cross-sectional area contains water (density $10^3 	ext{ kg m}^{-3}$). Initially the water level stands at $0.29 	ext{ m}$ from the bottom in each arm. Kerosene oil (a water-immiscible liquid) of density $800 	ext{ kg m}^{-3}$ is added to the left arm until its length is $0.1 	ext{ m}$, as shown in the schematic figure below. The ratio $\left(\frac{h_1}{h_2}ight)$ of the heights of the liquid in the two arms is-

Options:

(A) $\frac{15}{14}$

(B) $\frac{35}{33}$

(C) $\frac{7}{6}$

(D) $\frac{5}{4}$

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two Fluids

Imagine a classic U-tube experiment. Initially, we have pure water resting peacefully, perfectly balanced in both arms at a height of $0.29 \mathrm{m}$ from the bottom. The system is in perfect equilibrium.

Then, we disrupt this peace by pouring kerosene into the left arm until it forms a column of $0.1 \mathrm{m}$. Kerosene is lighter than water (density $800 	ext{ kg m}^{-3}$ compared to water's $1000 	ext{ kg m}^{-3}$) and immiscible, meaning it floats on top without mixing. The added weight of the kerosene pushes the water down in the left arm, forcing it to rise in the right arm. Our goal is to find the new ratio of the total liquid heights, $\frac{h_1}{h_2}$.

### The Unbreakable Rule: Conservation of Volume

Water is an incompressible fluid. This means the total volume of water in the U-tube cannot change. Because the U-tube has a uniform cross-sectional area, the total *length* of the water column must also remain constant.

Initially, the total length of the water column was $0.29 \mathrm{m} + 0.29 \mathrm{m} = 0.58 \mathrm{m}$.

In the final state, the total height of the liquid in the left arm is $h_1$. Since the top $0.1 \mathrm{m}$ is kerosene, the height of the water in the left arm is $(h_1 - 0.1)$. The height of the water in the right arm is simply $h_2$. Equating the final total water length to the initial total water length gives us our first master equation:

$$(h_1 - 0.1) + h_2 = 0.58$$

Rearranging this, we get:

$$h_1 + h_2 = 0.68 \quad \dots (1)$$

### Pascal's Law: Balancing the Pressures

Now, let's look at the physics of the fluids at rest. According to Pascal's Law, the pressure at any two points at the same horizontal level in a continuous, stationary fluid must be equal. The easiest place to apply this is at the very bottom of the U-tube.

The pressure at the bottom of the left arm is the sum of atmospheric pressure ($P_0$), the pressure exerted by the kerosene column, and the pressure exerted by the water column:

$$P_{	ext{left}} = P_0 + ho_k g (0.1) + ho_w g (h_1 - 0.1)$$

The pressure at the bottom of the right arm is the sum of atmospheric pressure and the pressure from its water column:

$$P_{	ext{right}} = P_0 + ho_w g h_2$$

Equating the two pressures ($P_{	ext{left}} = P_{	ext{right}}$), we notice that atmospheric pressure $P_0$ and the acceleration due to gravity $g$ cancel out beautifully:

$$ho_k (0.1) + ho_w (h_1 - 0.1) = ho_w h_2$$

Substituting the given densities ($ho_k = 800$ and $ho_w = 1000$):

$$800(0.1) + 1000(h_1 - 0.1) = 1000 h_2$$

$$80 + 1000 h_1 - 100 = 1000 h_2$$

$$1000(h_1 - h_2) = 20$$

$$h_1 - h_2 = 0.02 \quad \dots (2)$$

### The Final Calculation

We now have a beautifully simple system of two linear equations:
1. $h_1 + h_2 = 0.68$
2. $h_1 - h_2 = 0.02$

Adding the two equations eliminates $h_2$:

$$2h_1 = 0.70 \implies h_1 = 0.35 \mathrm{m}$$

Subtracting the second equation from the first eliminates $h_1$:

$$2h_2 = 0.66 \implies h_2 = 0.33 \mathrm{m}$$

Finally, we find the requested ratio:

$$\frac{h_1}{h_2} = \frac{0.35}{0.33} = \frac{35}{33}$$

**The correct option is (B).**

Select Answer:

Visualized Solution

$V_{initial} \to V_{final}$

$V_{water} = constant$

$h_{left} + h_{right} = constant$

$h_{1} + h_{2} = 0.68$

$P_{A} = P_{B}$

$P_{left} = P_{right}$

$ρ_{k} h_{k} + ρ_{w} h_{w 1} = ρ_{w} h_{w 2}$

$h_{1} - h_{2} = 0.02$

$h_{1}, h_{2} = ?$

$\frac{h _{1}}{h _{2}}$

$A_{1} \neq = A_{2}$

The Sigma Insight: Fluid Pressure and Pascal's Law

The Setup

The Unbreakable Rule

Pascal's Law

The Final Calculation

Similar Questions

There is a circular tube in a vertical plane. Two liquids which do not mix and of densities $d_{1}$ and $d_{2}$ are filled in the tube. Each liquid subtends $9 0^{\circ}$ angle at centre. Radius joining their interface makes an angle $α$ with vertical. Ratio $d_{1} / d_{2}$ is

There is a circular tube in a vertical plane. Two liquids which do not mix and of densities $d_{1}$ and $d_{2}$ are filled in the tube. Each liquid subtends $9 0^{\circ}$ angle at centre. Radius joining their interface makes an angle $α$ with vertical. Ratio $\frac{d _{1}}{d _{2}}$ is

A U-shaped tube contains a liquid of density $ρ$ and it is rotated about the line as shown in the figure. Find the difference in the levels of liquid column.

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Vinitial​→Vfinal​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Vwater​=constant

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } hleft​+hright​=constant

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h1​+h2​=0.68

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PA​=PB​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Pleft​=Pright​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ρk​hk​+ρw​hw1​=ρw​hw2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h1​−h2​=0.02

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h1​,h2​=?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h2​h1​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A1​=A2​