Question

A soap bubble is blown to a diameter of \( 7 \, \text{cm} \). \( 36960 \, \text{erg} \) of work is done in blowing it further. If the surface tension of the soap solution is \( 40 \, \text{dyne/cm} \), then the new radius is ______ \( \text{cm} \). Take: \( \left( \pi = \frac{22}{7} \right) \).

Answer 1

Correct Answer: 7

Work Done in Expanding a Soap Bubble:
 The work \( W \) done in increasing the surface area of a soap bubble is given by:
\[ W = \Delta U = S \Delta A \] where \( S \) is the surface tension and \( \Delta A \) is the increase in surface area.

Calculate Initial and Final Surface Areas:
Initial radius \( r = 7 \, \text{cm} \).
Initial surface area \( A_i = 4\pi r^2 = 4\pi (7)^2 \, \text{cm}^2 \).
Suppose the new radius is \( R \). Then the final surface area \( A_f \) is:
\[ A_f = 4\pi R^2 \]

Calculate the Change in Surface Area \( \Delta A \):
\[ \Delta A = A_f - A_i = 4\pi R^2 - 4\pi (7)^2 = 4\pi (R^2 - 49) \]

Use the Work Done to Solve for \( R \):
Given \( W = 36960 \, \text{erg} \) and \( S = 40 \, \text{dyne/cm} \), we have:
\[ 36960 = 40 \times 4\pi (R^2 - 49) \] 
Simplifying,
\[ 36960 = 160\pi (R^2 - 49) \] 
Using \( \pi = \frac{22}{7} \):
\[ 36960 \, \text{erg} = \frac{40 \, \text{dyne}}{\text{cm}} 8\pi \left[(R)^2 - \left(\frac{7}{2}\right)^2\right] \, \text{cm}^2 \]
\[ r = 7 \, \text{cm} \]

Conclusion:
The new radius of the soap bubble is \( 7 \, \text{cm} \).