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.

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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:

     \[
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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