Math Problem Statement

​please solve thus physics numerical

A U-shaped tube is partly filled with mercury. Then water is poured in its one limb and oil is poured in its other limb. Find the ratio of water and oil columns if the level of mercury is maintained the same in both limbs. Given densities are ρw = 1000 kg/m^3 for water and ρo = 900 kg/m^3 for oil.

Solution

The problem involves a U-shaped tube filled partially with mercury, with water poured into one limb and oil into the other. You are asked to find the ratio of the water and oil columns when the mercury level remains the same in both limbs.

Let's go step-by-step through the solution.

Given:

  • Density of water, ρw=1000kg/m3\rho_w = 1000 \, \text{kg/m}^3
  • Density of oil, ρo=900kg/m3\rho_o = 900 \, \text{kg/m}^3

Concept:

In this U-tube problem, the pressure at the same horizontal level in the mercury must be the same, so the pressure exerted by the water column in one limb must equal the pressure exerted by the oil column in the other limb.

Let:

  • hwh_w = height of the water column
  • hoh_o = height of the oil column

The pressure exerted by the water column at the mercury level is: Pw=ρwghwP_w = \rho_w g h_w The pressure exerted by the oil column at the same mercury level is: Po=ρoghoP_o = \rho_o g h_o

Since the pressures must be equal at the same level of mercury: ρwghw=ρogho\rho_w g h_w = \rho_o g h_o

We can cancel the gravitational acceleration gg from both sides: ρwhw=ρoho\rho_w h_w = \rho_o h_o

Now solve for the ratio of the heights of water and oil columns: hwho=ρoρw\frac{h_w}{h_o} = \frac{\rho_o}{\rho_w}

Substitute the given values: hwho=9001000=0.9\frac{h_w}{h_o} = \frac{900}{1000} = 0.9

Final Answer:

The ratio of the height of the water column to the height of the oil column is hwho=0.9\frac{h_w}{h_o} = 0.9.

This means that the water column will be 0.9 times the height of the oil column to maintain the same level of mercury in both limbs.


Do you have any questions or need more details on the explanation?

Here are 5 related questions for practice:

  1. How would the ratio change if the density of the oil were 800 kg/m³?
  2. What would happen if the oil density were greater than the water density?
  3. How does gravitational acceleration affect the result?
  4. What if the mercury level was not the same in both limbs—what would change?
  5. How can we measure the heights of these columns experimentally?

Tip: In a U-tube with different fluids, the pressures at the same horizontal level must always balance. This principle helps in many fluid mechanics problems.

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Math Problem Analysis

Mathematical Concepts

Fluid Mechanics
Pressure in Fluids
Density

Formulas

Pressure due to fluid column: P = ρgh
Equilibrium of fluid columns: ρw * h_w = ρo * h_o
Ratio of fluid columns heights: h_w/h_o = ρo/ρw

Theorems

Hydrostatic Pressure

Suitable Grade Level

High School