To solve this problem, we need to analyze how the addition of mercury affects the pressure in the manometer. The key here is to understand that the pressure exerted by a column of liquid is directly related to its height and density. In this case, we have mercury, which has a known density, and we can use the changes in height to find the pressure.

Understanding the Setup

We have a manometer with two arms filled with mercury. When we add an additional volume of mercury (10.0 cm³), it causes the mercury levels in both arms to change. Specifically, the left arm rises by 6.00 cm, and the right arm rises by 4.00 cm. The pressure in both arms is initially the same, denoted as p.

Calculating the Volume Change

First, let's calculate the total volume of mercury that has been added and how it affects the heights in each arm. The volume of mercury added is 10.0 cm³. Since the cross-sectional area of both arms is the same, we can relate the volume change to the height change using the formula:

• Volume = Area × Height

Let A be the cross-sectional area of the manometer. The volume increase in the left arm can be expressed as:

• Volume_left = A × 6.00 cm

Similarly, for the right arm:

• Volume_right = A × 4.00 cm

Setting Up the Equation

Since the total volume added is equal to the sum of the volumes in both arms, we can write:

• A × 6.00 cm + A × 4.00 cm = 10.0 cm³

Factoring out A gives us:

• A × (6.00 cm + 4.00 cm) = 10.0 cm³

This simplifies to:

• A × 10.00 cm = 10.0 cm³

From this, we can find the cross-sectional area:

• A = 1.0 cm²

Finding the Pressure p

Now that we have the area, we can relate the height changes to pressure. The pressure difference caused by the height of the mercury columns can be expressed using the hydrostatic pressure formula:

• ΔP = ρgh

Where:

• ΔP is the pressure difference
• ρ is the density of mercury (approximately 13,600 kg/m³)
• g is the acceleration due to gravity (approximately 9.81 m/s²)
• h is the height difference in meters

In our case, the height difference between the two arms after adding the mercury is:

• h = 6.00 cm - 4.00 cm = 2.00 cm = 0.02 m

Now we can calculate the pressure difference:

• ΔP = 13,600 kg/m³ × 9.81 m/s² × 0.02 m

Calculating this gives:

• ΔP ≈ 2,673.6 Pa

Since the pressure in both arms was initially the same, the pressure p can be expressed as:

• p = p₀ + ΔP

Assuming p₀ (the atmospheric pressure) is approximately 101,325 Pa, we find:

• p = 101,325 Pa + 2,673.6 Pa ≈ 103,998.6 Pa

Final Result

Thus, the pressure p in the manometer after admitting the additional mercury is approximately:

• p ≈ 104,000 Pa

This result illustrates how the addition of a specific volume of liquid can affect pressure measurements in a manometer, highlighting the relationship between liquid height and pressure in fluid mechanics.