Answer :
To solve the problem of finding the density of another fluid in the barometer, we can use the concept of pressure equilibrium. Here's a step-by-step approach:
1. Understanding the Principle:
- In a barometer, the pressure exerted by the mercury column is equal to the pressure exerted by the other fluid column when they are in equilibrium and open to the same atmospheric pressure.
- The pressure exerted by a fluid column is given by the product of its height, density, and gravitational acceleration (Pressure = Height x Density x g), where g is a constant (acceleration due to gravity).
2. Given Data:
- Height of mercury column = 69 cm
- Height of the other fluid column = 176 cm
- Density of mercury = 13.6 g/cm³ (this is a known value and is consistent for mercury)
3. Using Pressure Equilibrium:
- Since the pressures are equal, we have:
[tex]\[
\text{Pressure by mercury} = \text{Pressure by the other fluid}
\][/tex]
- Writing in terms of height and density:
[tex]\[
\text{Height of mercury} \times \text{Density of mercury} = \text{Height of fluid} \times \text{Density of fluid}
\][/tex]
4. Calculating Density of the Other Fluid:
- Rearranging the formula to solve for the density of the other fluid:
[tex]\[
\text{Density of fluid} = \frac{\text{Height of mercury} \times \text{Density of mercury}}{\text{Height of fluid}}
\][/tex]
- Plugging in the given values:
[tex]\[
\text{Density of fluid} = \frac{69 \, \text{cm} \times 13.6 \, \text{g/cm}^3}{176 \, \text{cm}}
\][/tex]
- Simplifying this expression gives us the density of the fluid:
[tex]\[
\text{Density of fluid} \approx 5.33 \, \text{g/cm}^3
\][/tex]
So, the density of the other fluid in the barometer is approximately 5.33 g/cm³.
1. Understanding the Principle:
- In a barometer, the pressure exerted by the mercury column is equal to the pressure exerted by the other fluid column when they are in equilibrium and open to the same atmospheric pressure.
- The pressure exerted by a fluid column is given by the product of its height, density, and gravitational acceleration (Pressure = Height x Density x g), where g is a constant (acceleration due to gravity).
2. Given Data:
- Height of mercury column = 69 cm
- Height of the other fluid column = 176 cm
- Density of mercury = 13.6 g/cm³ (this is a known value and is consistent for mercury)
3. Using Pressure Equilibrium:
- Since the pressures are equal, we have:
[tex]\[
\text{Pressure by mercury} = \text{Pressure by the other fluid}
\][/tex]
- Writing in terms of height and density:
[tex]\[
\text{Height of mercury} \times \text{Density of mercury} = \text{Height of fluid} \times \text{Density of fluid}
\][/tex]
4. Calculating Density of the Other Fluid:
- Rearranging the formula to solve for the density of the other fluid:
[tex]\[
\text{Density of fluid} = \frac{\text{Height of mercury} \times \text{Density of mercury}}{\text{Height of fluid}}
\][/tex]
- Plugging in the given values:
[tex]\[
\text{Density of fluid} = \frac{69 \, \text{cm} \times 13.6 \, \text{g/cm}^3}{176 \, \text{cm}}
\][/tex]
- Simplifying this expression gives us the density of the fluid:
[tex]\[
\text{Density of fluid} \approx 5.33 \, \text{g/cm}^3
\][/tex]
So, the density of the other fluid in the barometer is approximately 5.33 g/cm³.
