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. 4.2 cm
• B. 5.6 cm
• C. 2.8 cm
• D. 1.4 cm

Hint:

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.

Answer 1

The Correct Option is C

Step 1: Picture what holds the water out.
Water tries to push into the small hole because of the pressure of the water column outside. A curved water surface (meniscus) forms in the hole, and its surface tension creates an extra pressure that pushes back. The vessel can go down until these two just balance.
Step 2: Write the balance condition.
The outside water pressure at depth $h$ is $h\rho g$. The extra pressure from the curved surface in a hole of radius $r$ is $\dfrac{2T}{r}$. Set them equal: \[ h\rho g = \frac{2T}{r} \]
Step 3: List the given values.
Radius $r = 0.5$ mm $= 5\times10^{-4}$ m, surface tension $T = 7\times10^{-2}\ Nm^{-1}$, density $\rho = 1000\ \text{kg/m}^{3}$, gravity $g = 10\ ms^{-2}$.
Step 4: Put values into the equation.
\[ h(1000)(10) = \frac{2 \times 7\times10^{-2}}{5\times10^{-4}} \]
Step 5: Work out the right side.
\[ h\times10^{4} = \frac{14\times10^{-2}}{5\times10^{-4}} = 280 \]
Step 6: Solve for the depth.
\[ h = \frac{280}{10^{4}} = 0.028\ \text{m} = 2.8\ \text{cm} \] \[ \boxed{h = 2.8\ \text{cm}} \]