Question

A soap bubble, having radius of \(1 mm\), is blown from a detergent solution having a surface tension of \(2.5 \times 10^{-2} \)\(\frac Nm\). The pressure inside the bubble equals at a point \(Z_0\) below the free surface of water in a container. Taking \(g=10 m/s^2,\) density of water-\(103 kg/m^3\), the value of \(Z_0\) is: 

• A. \(100 cm\)
• B. \(10 cm\)
• C. \(1 cm\)
• D. \(0.5 cm\)

Answer 1

The Correct Option is C

To solve for the distance \(Z_0\) below the free surface of water where the pressure equals the pressure inside a soap bubble, we need to consider the formula for excess pressure inside a soap bubble and equate it to the hydrostatic pressure at depth \(Z_0\).

The excess pressure inside a soap bubble is given by:

\( \Delta P = \frac{4T}{R} \)

where \(T\) is the surface tension of the soap solution and \(R\) is the radius of the bubble. In the given problem:

• \(T = 2.5 \times 10^{-2} \, \frac{N}{m}\)
• \(R = 1 \, mm = 1 \times 10^{-3} \, m\)

Substituting these values, we find:

\( \Delta P = \frac{4 \times 2.5 \times 10^{-2}}{1 \times 10^{-3}} = 100 \, \frac{N}{m^2} \)

Now, equate this excess pressure to the hydrostatic pressure at depth \(Z_0\) in water:

\( P = \rho g Z_0 \)

where \(\rho = 103 \, kg/m^3\) is the density of water and \(g = 10 \, m/s^2\).

Thus, \(\rho g Z_0 = 100\)

Substituting the given values:

\( 103 \times 10 \times Z_0 = 100 \)

Solving for \(Z_0\) gives:

\( Z_0 = \frac{100}{1030} \approx 0.097 \, m = 1 \, cm \)

Therefore, the correct value of \(Z_0\) is \(1 \, cm\). The correct answer is \(1 \, cm\).