Question

A soap bubble of surface tension \(0.04\,\text{N/m}\) is blown to a diameter of \(7\,\text{cm}\). If \((15000 - x)\,\mu\text{J}\) of work is done in blowing it further to make its diameter \(14\,\text{cm}\) \((\pi = 22/7)\), then the value of \(x\) is ________.

Hint:

Soap bubbles have two surfaces, hence energy involves a factor of 2.

Answer 1

Correct Answer: 11304

To solve the problem, we need to calculate the work done in blowing a soap bubble from an initial diameter of \(7\,\text{cm}\) to a final diameter of \(14\,\text{cm}\). The work done \(W\) is given by the change in surface energy, which is determined by the surface tension \(\gamma\) and the change in surface area of the bubble.
First, calculate the initial and final surface areas of the bubble. The formula for the surface area \(A\) of a sphere is:
\(A = 4\pi r^2\), where \(r\) is the radius.
The initial radius \(r_i = \frac{7}{2} = 3.5\,\text{cm}\). The initial surface area \(A_i\) is:
\(A_i = 4\pi (3.5\,\text{cm})^2 = 4 \times \frac{22}{7} \times 12.25\,\text{cm}^2 = 154\,\text{cm}^2\).
The final radius \(r_f = \frac{14}{2} = 7\,\text{cm}\). The final surface area \(A_f\) is:
\(A_f = 4\pi (7\,\text{cm})^2 = 4 \times \frac{22}{7} \times 49\,\text{cm}^2 = 616\,\text{cm}^2\).
The change in surface area \(\Delta A = A_f - A_i = 616 - 154 = 462\,\text{cm}^2\).
The surface tension \(\gamma = 0.04\,\text{N/m} = 0.04\,\text{J/m}^2\) (since \(1\,\text{N} = 1\,\text{J/m}\)). The work done \(W\) in joules is:
\(W = \gamma \cdot 2 \cdot \Delta A\), where the factor of 2 accounts for the inner and outer surfaces of the bubble.
\(W = 0.04 \cdot 2 \cdot 462 \times 10^{-4}\,\text{m}^2 = 0.04 \cdot 2 \cdot 0.0462\,\text{m}^2 = 0.003696\,\text{J}\).
Convert joules to microjoules: \(0.003696\,\text{J} = 3696\,\mu\text{J}\).
Given \(15000 - x = 3696\), solve for \(x\):
\(x = 15000 - 3696 = 11304\).
Thus, the value of \(x\) is confirmed as \(11304\), which falls within the expected range of 11304 to 11304.