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Q10.10:
In the previous problem, if 15.0 cm of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms ? (Specific gravity of mercury = 13.6)

Solution :

Step 1: Identifying the Physical Principle

The problem relies on the principle of hydrostatic equilibrium in a U-tube. For a fluid in equilibrium, the pressure at any two points at the same horizontal level in the same continuous liquid must be equal [Per Pascal’s Law of Hydrostatics: $P = P_0 + \rho gh$].

Step 2: Defining Given Variables

  • Density of water ($\rho_w$): $1.0 \, \text{g/cm}^3$
  • Density of spirit ($\rho_s$): $0.8 \, \text{g/cm}^3$
  • Density of mercury ($\rho_m$): $13.6 \, \text{g/cm}^3$
  • Height of water column ($h_w$): $15.0 \, \text{cm}$
  • Height of spirit column ($h_s$): $15.0 \, \text{cm}$

Step 3: Analytical Derivation of the Level Difference

Let the initial height of mercury be equal in both arms. When water is added to one arm and spirit to the other, the mercury level shifts. Let the mercury in the water-arm side drop by $x$ and rise by $x$ in the spirit-arm side. The total difference in mercury levels becomes $2x$.

Pressure balance at the interface of mercury and the other liquids:

$P_{\text{right}} = P_{\text{left}}$

$\rho_w g h_w = \rho_s g h_s + \rho_m g (2x)$

Since $g$ is common to all terms, we simplify:

$\rho_w h_w = \rho_s h_s + \rho_m (2x)$

Step 4: Algebraic Substitution & Calculation

Substitute the known values into the equation:

$(1.0)(15.0) = (0.8)(15.0) + (13.6)(2x)$

$15.0 = 12.0 + 27.2x$

$15.0 - 12.0 = 27.2x$

$3.0 = 27.2x$

$x = \frac{3.0}{27.2} \approx 0.1103 \, \text{cm}$

The difference in the levels of mercury is $2x$:

Difference $= 2 \times 0.1103 = 0.2206 \, \text{cm}$

Step 5: Visual Representation

2x Water Spirit

Final Solution: The difference in the levels of mercury in the two arms is 0.221 cm.


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