Problem 69 - Master Resource Book in JEE Main Physics

Question

The area of cross-section of one limb of an U-tube is twice that of the other. Both the limbs contains mercury at the same level. Water is poured in the wider tube so that mercury level in it goes down by \(1 \mathrm{~cm}\). The height of water column is (Density of water \(=10^{3} \mathrm{kgm}^{-\text {I }}\), density of mercury \(=13.6 \times 10^{3} \mathrm{kgm}^{-3}\) ) \(\quad\) ?erala CET 200s] (a) \(13.6 \mathrm{~cm}\) (b) \(40.8 \mathrm{~cm}\) (c) \(6.8 \mathrm{~cm}\) (d) \(54.4 \mathrm{~cm}\)

Step-by-Step Solution

Answer

The height of the water column is 40.8 cm (option b).

Step 1: Understand the Problem

We are given that the cross-sectional area of one limb in a U-tube is twice that of the other, with equal mercury levels initially. Water is poured into the wider tube, causing the mercury level to drop by 1 cm. Our goal is to find the height of the water column that causes this change, using the density of water and mercury.

Step 2: Define Area Relationship

Let the cross-sectional area of the narrower limb be \(A\) and the wider limb be \(2A\). The relationship between the changes in mercury levels in the limbs due to water addition must now be described.

Step 3: Determine Mercury Movement

When the mercury level falls by 1 cm in the wider limb, it rises in the narrower limb. Because the volume of mercury displaced is equal in both limbs, if the mercury drops by \(1\) cm in the wider limb, it must rise by \(x\) cm in the narrower limb, where \(2A \times 1 = A \times x\). Thus, \(x = 2\) cm.

Step 4: Apply Hydrostatic Principle

Because the mercury column rises by 2 cm in the narrower limb and falls by 1 cm in the wider limb, the difference in mercury levels becomes 3 cm. The hydrostatic pressure balance states that the pressure from the water column equals the pressure due to the 3 cm difference in mercury: \(h_{w} \times \rho_{w} = 3 \times \rho_m \), where \(\rho_w\) is the density of water and \(\rho_m\) is the density of mercury.

Step 5: Solve for Water Column Height

Using \(h_w \times 1000 = 3 \times 13600\), solve for \(h_w\). This gives \(h_w = \frac{3 \times 13600}{1000} = 40.8\) cm.

Key Concepts

Density of Liquids

In the world of hydrostatics, understanding the density of liquids is crucial. Density is defined as mass per unit volume and is symbolically presented as \( \rho \). For water and mercury, densities are especially important in the given exercise. When you compare these densities, you notice that mercury is significantly denser than water. Water has a density of \( 1000 \, \text{kg/m}^3 \) while mercury's density is \( 13600 \, \text{kg/m}^3 \).
This large difference in density is why a small change in the mercury level corresponds to a much larger column of water being able to balance it. The fact that mercury is so dense means it doesn't rise or fall as much as less dense liquids, like water, when subjected to the same changes in pressure.
Being dense means that for a given pressure, a shorter column of mercury is needed compared to a column of a lighter liquid, like water, which needs to be taller to exert the same pressure.

Pressure Balance

Pressure balance plays a central role in solving U-tube problems. The pressure exerted by liquids in a container at rest in a stable temperature environment is derived from hydrostatics. When a liquid is added to a side of the U-tube, it creates additional pressure due to its height and density, expressed as \( P = h \rho g \).
In this exercise, pouring water into the wider limb of the U-tube alters the mercury levels because water has exerted pressure enough to push the mercury down.
The critical concept here is that the pressure in both limbs must balance out. That means the pressure from the column of water in the wider limb must be equal to the pressure difference generated in the mercury. This is expressed in the relation: \( h_w \rho_w = 3 \rho_m \), where the difference of 3 cm accounts for mercury levels adjusting across both limbs.

U-tube Problem

The U-tube problem serves as a fundamental demonstration of pressure equilibrium in interconnected columns of liquid. In a classic U-tube, pressures at points of equal height are balanced, which is a crucial understanding in this exercise.
Initially, the mercury levels in both limbs are at the same height. Pouring water in the wider limb disturbs this balance, causing the mercury level there to drop by 1 cm.
By understanding this, you can see how the movement in one limb affects the other. When dealing with a U-tube, equilibrium is disturbed by added pressure, but eventually restored by the movement of liquids, accurately predicted using principles of hydrostatics.

Cross-sectional Area Relationships

The cross-sectional area of the U-tube limbs directly impacts how liquids move between the two sides when disturbed. In this scenario, the wider limb has a cross-sectional area of \( 2A \), while the narrower limb has an area of \( A \).
This relationship is crucial in determining the extent of mercury's movement. When the level in the wider limb drops by 1 cm, due to its greater area, an equivalent volume in the narrow limb rises by 2 cm, maintaining the volume conservation of mercury.
By calculating this rise, one can establish how much water must be added to induce this change in equilibrium. The cross-sectional area differences essentially dictate the ratio of movement between the two limbs.