Question:
A long cylindrical vessel of glass having a hole of 0.5 mm radius at its bottom is slowly lowered vertically into a deep water bath. If the surface tension of water is \( 7 \times 10^{-2} \, Nm^{-1} \) then the maximum depth the vessel can be submerged without water entering through the hole is: (Acceleration due to gravity \( = 10 \, ms^{-2} \))
A long cylindrical vessel of glass having a hole of 0.5 mm radius at its bottom is slowly lowered vertically into a deep water bath. If the surface tension of water is \( 7 \times 10^{-2} \, Nm^{-1} \) then the maximum depth the vessel can be submerged without water entering through the hole is: (Acceleration due to gravity \( = 10 \, ms^{-2} \))
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Always convert all given measurements into standard SI units (meters, kilograms, seconds) before substituting them into formulas. This simple step prevents order-of-magnitude errors in your final answer.
Updated On: Jun 7, 2026
- 4.2 cm
- 5.6 cm
- 2.8 cm
- 1.4 cm
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The Correct Option is C
Solution and Explanation
Concept:
For water to stay out of the cylindrical vessel, the excess pressure created inside the water meniscus at the small hole must balance the external hydrostatic pressure from the water column:
\[
h \cdot \rho \cdot g = \frac{2T}{r}
\]
Step 1: Substituting given parameters into the pressure balance equation.
Given parameters:
• Radius of the hole \( r = 0.5 \, \text{mm} = 5 \times 10^{-4} \, \text{m} \)
• Surface tension \( T = 7 \times 10^{-2} \, Nm^{-1} \)
• Density of water \( \rho = 1000 \, \text{kg/m}^3 \)
• Gravity \( g = 10 \, ms^{-2} \)
Step 2: Isolating depth parameter \( h \).
\[ h (1000)(10) = \frac{2 \times 7 \times 10^{-2}}{5 \times 10^{-4}} \] \[ h \times 10^4 = \frac{14 \times 10^{-2}}{5 \times 10^{-4}} = 2.8 \times 10^2 = 280 \] \[ h = \frac{280}{10^4} = 0.028 \, \text{m} \]
Step 3: Converting to centimeters.
\[ h = 0.028 \times 100 \, \text{cm} = 2.8 \, \text{cm} \] This matches option (C) perfectly.
Step 1: Substituting given parameters into the pressure balance equation.
Given parameters:
• Radius of the hole \( r = 0.5 \, \text{mm} = 5 \times 10^{-4} \, \text{m} \)
• Surface tension \( T = 7 \times 10^{-2} \, Nm^{-1} \)
• Density of water \( \rho = 1000 \, \text{kg/m}^3 \)
• Gravity \( g = 10 \, ms^{-2} \)
Step 2: Isolating depth parameter \( h \).
\[ h (1000)(10) = \frac{2 \times 7 \times 10^{-2}}{5 \times 10^{-4}} \] \[ h \times 10^4 = \frac{14 \times 10^{-2}}{5 \times 10^{-4}} = 2.8 \times 10^2 = 280 \] \[ h = \frac{280}{10^4} = 0.028 \, \text{m} \]
Step 3: Converting to centimeters.
\[ h = 0.028 \times 100 \, \text{cm} = 2.8 \, \text{cm} \] This matches option (C) perfectly.
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