10. The manometer below (picture A) holds a thick immiscible fluid. When a 22-cm tall column of water in poured into the tube (picture B), the equilibrium value of H is 16.5 cm. Calculate the density of the immiscible fluid.
10. The manometer below (picture A) holds a thick immiscible fluid. When a 22-cm tall column of water in poured into the tube (picture B), the equilibrium value of H is 16.5 cm. Calculate the density of the immiscible fluid.
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![### Manometer Fluid Density Calculation
**Problem Statement:**
10. The manometer below (picture A) holds a thick immiscible fluid. When a 22-cm tall column of water is poured into the tube (picture B), the equilibrium value of H is 16.5 cm. Calculate the density of the immiscible fluid.
#### Illustrations:
**Picture (A):**
- Displays a U-shaped manometer which contains a thick immiscible fluid. The fluid appears to be at the same level in both arms of the U-tube, indicating that there is no additional pressure applied to either side.
**Picture (B):**
- Shows the same U-shaped manometer where a 22-cm tall column of water has been added to the left arm of the U-tube.
- The increased pressure due to the water column has displaced the immiscible fluid downward in the left arm and upward in the right arm.
- The height difference, H, between the two arms of the immiscible fluid, is shown to be 16.5 cm.
#### Calculation Explanation:
1. **Given Data:**
- Height of water column (\( h_{water} \)): 22 cm
- Height difference of immiscible fluid (\( H \)): 16.5 cm
- Density of water (\( \rho_{water} \)): \( 1 \text{ g/cm}^3 \) or \( 1000 \text{ kg/m}^3 \)
2. **Objective:**
- To calculate the density of the immiscible fluid (\( \rho_{immiscible} \)).
3. **Formula and Calculation:**
- The pressure due to the column of water is given by: \( P_{water} = \rho_{water} \cdot g \cdot h_{water} \).
- The pressure due to the immiscible fluid on the other side is given by: \( P_{immiscible} = \rho_{immiscible} \cdot g \cdot H \).
- At equilibrium, these pressures must be equal, hence:
\[
\rho_{water} \cdot h_{water} = \rho_{immiscible} \cdot H
\]
- Rearranging to solve for the density of the immiscible fluid:
\[](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5498129d-fa90-4054-9767-1cffe54599e5%2F28963a21-1574-481d-b443-42dbf2f3cf2d%2F9gosu58_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Manometer Fluid Density Calculation
**Problem Statement:**
10. The manometer below (picture A) holds a thick immiscible fluid. When a 22-cm tall column of water is poured into the tube (picture B), the equilibrium value of H is 16.5 cm. Calculate the density of the immiscible fluid.
#### Illustrations:
**Picture (A):**
- Displays a U-shaped manometer which contains a thick immiscible fluid. The fluid appears to be at the same level in both arms of the U-tube, indicating that there is no additional pressure applied to either side.
**Picture (B):**
- Shows the same U-shaped manometer where a 22-cm tall column of water has been added to the left arm of the U-tube.
- The increased pressure due to the water column has displaced the immiscible fluid downward in the left arm and upward in the right arm.
- The height difference, H, between the two arms of the immiscible fluid, is shown to be 16.5 cm.
#### Calculation Explanation:
1. **Given Data:**
- Height of water column (\( h_{water} \)): 22 cm
- Height difference of immiscible fluid (\( H \)): 16.5 cm
- Density of water (\( \rho_{water} \)): \( 1 \text{ g/cm}^3 \) or \( 1000 \text{ kg/m}^3 \)
2. **Objective:**
- To calculate the density of the immiscible fluid (\( \rho_{immiscible} \)).
3. **Formula and Calculation:**
- The pressure due to the column of water is given by: \( P_{water} = \rho_{water} \cdot g \cdot h_{water} \).
- The pressure due to the immiscible fluid on the other side is given by: \( P_{immiscible} = \rho_{immiscible} \cdot g \cdot H \).
- At equilibrium, these pressures must be equal, hence:
\[
\rho_{water} \cdot h_{water} = \rho_{immiscible} \cdot H
\]
- Rearranging to solve for the density of the immiscible fluid:
\[
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