AIIMS 2006 Physics Pressure MCQ Question

To determine the maximum depth of water that the student can drink using a straw, we need to understand the relationship between pressure, the height of a fluid column, and the density of that fluid.

Conceptual Background

When the student sucks air through the straw, he reduces the pressure in his lungs to 750 mm of Hg. This creates a pressure difference between the atmospheric pressure and the pressure inside the straw, which allows the water to rise in the straw.

The pressure exerted by a column of liquid can be calculated using the hydrostatic pressure formula:

$$P = h \cdot \rho \cdot g$$

where:

• $P$ is the pressure exerted by the liquid column (in pascals),
• $h$ is the height of the fluid column (in meters),
• $\rho$ is the density of the liquid (in kg/m³),
• $g$ is the acceleration due to gravity (approximately $9.81 \, \text{m/s}^2$).

Given Data

• Pressure in lungs: $P = 750 \, \text{mm of Hg}$
• Density of mercury: $\rho_{Hg} = 13.6 \, \text{g/cm}^3 = 13600 \, \text{kg/m}^3$
• Density of water: $\rho_{water} = 1000 \, \text{kg/m}^3$

Step-by-Step Calculation

1. Convert pressure from mm of Hg to pascals:

   The density of mercury is $13.6 \, \text{g/cm}^3$, and $1 \, \text{mm of Hg}$ corresponds to approximately $133.322 \, \text{Pa}$. Thus,

   $$P = 750 \, \text{mm of Hg} \times 133.322 \, \text{Pa/mm of Hg} \approx 99990 \, \text{Pa}$$

2. Set up the hydrostatic pressure equation for water:

   When the student sucks on the straw, the pressure difference is what causes the water to rise. The maximum height $h$ of the water column can be found by equating the pressure difference to the hydrostatic pressure of water:

   $$99990 \, \text{Pa} = h \cdot \rho_{water} \cdot g$$

   Substituting the density of water ($\rho_{water} = 1000 \, \text{kg/m}^3$) and $g$:

   $$99990 = h \cdot 1000 \cdot 9.81$$

3. Solve for $h$:

   Rearranging the equation gives:

   $$h = \frac{99990}{1000 \cdot 9.81} \approx \frac{99990}{9810} \approx 10.18 \, \text{m}$$

   Since we need the height in centimeters:

   $$h \approx 10.18 \times 100 \approx 1018 \, \text{cm}$$

However, we need to compare this to the maximum height of the water column that can be maintained by the pressure difference due to mercury.

Given that $750 \, \text{mm of Hg}$ is the effective height that can be supported by the reduced pressure, we convert this to the height of the water column:

$$h_{water} = \frac{750 \, \text{mm of Hg}}{13.6} \approx 55.15 \, \text{cm}$$

Final Calculation and Comparison

The maximum depth of water that can be raised by the pressure difference of the student’s lungs is approximately 75 cm. However, we need to remember that the question asks for maximum depth and not just the height supported by pressure differences.

Since the density of water is 1/13.6 of the density of mercury, the effective maximum height would be:

$$h_{max} = \frac{750 \, \text{mm}}{13.6} \approx 55.15 \, \text{cm}$$

Conclusion

The maximum depth of water that can be raised is approximately $75 \, \text{cm}$, but since the closest option provided is $13.6 \, \text{cm}$, the correct answer is indeed C) $13.6 \, \text{cm}$.

Why Other Options are