A U-tube of uniform cross-section is partially filled with a liquid I. Another liquid II which does

Q: Why didn't we use the 2 cm rise in our calculations?

Because the problem specifies that the final liquid levels in both arms are equal. This directly dictates that the heights of both columns above the interface are identical, making the 2 cm value redundant for finding the density.

Q: Can we equate pressures at any horizontal level?

Only if the points lie within the same continuous, stationary fluid. That's why we chose the interface level, as the fluid below this level is entirely Liquid I.

Q: What is specific gravity?

Specific gravity is the ratio of the density of a substance to the density of a reference substance (usually water at 4°C). Since it's a ratio, equal densities mean equal specific gravities.

Q: What would happen if the cross-section of the U-tube was not uniform?

The heights of the columns above the interface would still be equal if the top levels were equal, so the specific gravity would still be 1.1. However, the 2 cm rise on one side would correspond to a different fall on the other side.

Q: Is the pressure at the interface always equal to the pressure at the same level in the other arm?

Yes, as long as the fluid below that level is continuous and stationary, ensuring hydrostatic equilibrium.

Question

A U-tube of uniform cross-section is partially filled with a liquid I. Another liquid II which does not mix with liquid I is poured into one side. It is found that the liquid levels of the two sides of the tube are the same, while the level of liquid I has risen by $2	ext{ cm}$ on the other side. If the specific gravity of liquid I is $1.1$, the specific gravity of liquid II must be

Options:

(A) 1.12

(B) 1.1

(C) 1.05

(D) 1.0

Sigma Percentile · Accepted Answer

### The Deceptive Simplicity of Hydrostatics

Many physics problems in competitive exams like the **JEE** are designed to test your conceptual clarity rather than your algebraic stamina. This classic 1983 problem is a prime example. At first glance, the mention of a **$2	ext{ cm}$ rise** in the liquid level might tempt you to set up complex volume conservation equations. However, if we step back and look at the fundamental principles of fluid statics, the solution unfolds with elegant simplicity.

Let's dive deep into the physics of this system and understand why some information is beautifully redundant.

---

### Setting Up the Visual Stage

Imagine a symmetric U-tube of uniform cross-section. Initially, it contains only **Liquid I**. When we pour **Liquid II** (which is immiscible with Liquid I) into the left arm, it exerts pressure and pushes the column of Liquid I downwards on the left side. Consequently, Liquid I rises in the right arm.

Once the system reaches static equilibrium, we observe two key facts:
1. The top surfaces of the liquids in both arms are at the **exact same horizontal level**.
2. There is a clear boundary, or **interface**, between Liquid II and Liquid I in the left arm.

Let's draw a horizontal reference line passing through this interface on the left side. Let this be point $1$. On the right side, at the exact same horizontal level, let's mark point $2$.

$$
	ext{Reference Level: } y = y_{	ext{interface}}
$$

---

### The Magic of Pascal's Principle

In a stationary, continuous fluid, the pressure at any two points lying on the same horizontal plane must be equal. Since the fluid below our reference line is entirely **Liquid I** and is continuous from the left arm to the right arm, we can confidently write:

$$P_1 = P_2$$

Now, let's express the hydrostatic pressure at both points. Let $P_0$ be the atmospheric pressure acting on the open tops of both arms.

* **At Point 1 (Left Arm):** The pressure is due to the atmosphere plus the column of Liquid II of height $h$ above the interface:

$$P_1 = P_0 + ho_{II} g h$$

* **At Point 2 (Right Arm):** The pressure is due to the atmosphere plus the column of Liquid I of height $h$ above the reference level:

$$P_2 = P_0 + ho_I g h$$

Because the top surfaces of both arms are at the same horizontal level, the height $h$ of Liquid II in the left arm is **exactly equal** to the height of Liquid I above the reference level in the right arm!

---

### The Elegant Cancellation

Equating the two pressures:

$$P_0 + ho_{II} g h = P_0 + ho_I g h$$

Subtracting $P_0$ from both sides:

$$ho_{II} g h = ho_I g h$$

Dividing both sides by $g h$:

$$ho_{II} = ho_I$$

This simple result tells us that for the top levels to remain equal, the densities of the two liquids must be **identical**!

Since specific gravity is directly proportional to density, we have:

$$	ext{Specific Gravity of Liquid II} = 	ext{Specific Gravity of Liquid I} = 1.1$$

Thus, the correct option is **(b)**.

---

### Why the "2 cm" is a Clever Distractor

The information that Liquid I rose by $2	ext{ cm}$ is a physical consequence of pouring Liquid II, but it is completely unnecessary for finding the specific gravity. It simply describes the transition from the initial state to the final equilibrium state. This is a classic trap to watch out for in JEE—always identify which physical constraints are active and which details are merely descriptive!

Select Answer:

Visualized Solution

Visualizing the U-Tube System

Hydrostatic Pressure & Pascal's Principle

Identifying Column Heights

Expressing Pressure at Points $1$ and $2$

Applying Equilibrium Condition

Simplifying the Equation

Determining Specific Gravity

Why the '2 cm' Rise is Redundant

The Sigma Insight: Fluid Pressure and Pascal's Law

The Deceptive Simplicity of Hydrostatics

Setting Up the Visual Stage

The Magic of Pascal's Principle

The Elegant Cancellation

Why the "2 cm" is a Clever Distractor

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