The Question is from neet-physics mechanical-properties-of-fluids variation-of-pressure-upthrust, and it was created by the author ABDUL JILANI BASHA

Question

A U-tube contains two immiscible liquids of densities $\rho_1$ and $\rho_2$ ($\rho_2 > \rho_1$). The heights of the liquid columns above the common interface are $h_1$ and $h_2$ respectively. If a small pressure difference $\Delta P$ is applied to one arm, the interface level shifts by $\Delta h$. Which expression gives $\Delta h$?

ABDUL JILANI BASHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
 **Primary Concept**: Hydrostatic pressure and pressure difference in fluids.
 **Related Topics**: Pascal's law, fluid statics.
 **Prerequisites**: Understanding of pressure, density, and hydrostatic pressure.
## SOLUTION PATHWAY
**Given Information**:
	Two immiscible liquids with densities $ho_1$ and $ho_2$ ($ho_2 > ho_1$)
	Heights of liquid columns: $h_1$ and $h_2$
	Pressure difference applied: $\Delta P$
	Interface level shift: $\Delta h$
**Method & Steps**:
1. Consider the pressure balance before the pressure difference is applied. At the interface, the pressure is the same in both arms.
2. When $\Delta P$ is applied, the pressure in one arm increases by $\Delta P$. This causes the interface to shift by $\Delta h$.
3. The pressure difference $\Delta P$ is balanced by the change in hydrostatic pressure due to the shift in the interface.  The change in height of the liquid with density $ho_2$ is $-\Delta h$, and the change in height of the liquid with density $ho_1$ is $+\Delta h$.
4. The pressure difference is given by:
$\Delta P = ho_2 g (-\Delta h) - ho_1 g (\Delta h) = -\Delta h g (ho_2 - ho_1)$
5. Solving for $\Delta h$:
$\Delta h = \frac{\Delta P}{g(ho_2 - ho_1)}$

Result: $\Delta h = \frac{\Delta P}{g(ho_2 - ho_1)}$
 Verification: The units are consistent (Pascals / (m/s² * kg/m³) = meters).
## OPTION ANALYSIS
[A] ✓: Correct. This is the derived expression for the change in interface height.
[B] :x: Incorrect.  This would be the case if the pressure difference was distributed evenly across both liquids, which is not true.
[C] :x: Incorrect. This expression involves the initial heights, which are irrelevant to the change in interface height due to a small pressure difference.
[D] :x: Incorrect. This considers only the denser liquid, ignoring the effect of the less dense liquid.
## LEARNING AIDS
 **Common Mistakes**:
  1.Incorrectly applying the pressure difference:  Forgetting that the pressure difference affects both liquids, leading to a change in the relative heights.
   How to avoid: Draw a clear diagram before and after applying the pressure difference, carefully considering the changes in the heights of both columns.
  2.Ignoring the sign convention: Neglecting that the height changes are in opposite directions for the two liquids.
   How to avoid:  Establish a clear coordinate system and carefully track the sign of the height changes.
 **Quick Tips**:
  1.Remember that the pressure difference is balanced by the difference in hydrostatic pressure between the two liquids.
  2.Use a diagram to visualize the height changes before and after applying the pressure difference.

ABDUL JILANI BASHA · Answer

The correct Option is "$\frac{\Delta P}{g(\rho_2 - \rho_1)}$"

ABDUL JILANI BASHA · Answer

$\frac{\Delta P}{g(\rho_2 + \rho_1)}$

ABDUL JILANI BASHA · Answer

$\frac{\Delta P h_1 h_2}{g(h_2 \rho_2 - h_1 \rho_1)}$

ABDUL JILANI BASHA · Answer

$\frac{\Delta P}{g\rho_2}$

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