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 \( W \) required to increase the surface area of a soap bubble is expressed as: \( W = \Delta U = S \Delta A \), where \( S \) denotes surface tension and \( \Delta A \) represents the change in surface area.

Calculate Initial and Final Surface Areas:
The initial radius is given as \( r = 7 \, \text{cm} \).
The initial surface area is \( A_i = 4\pi r^2 = 4\pi (7)^2 \, \text{cm}^2 \).
Assuming a new radius \( R \), the final surface area \( A_f \) is calculated as: \( A_f = 4\pi R^2 \).

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 \( S = 40 \, \text{dyne/cm} \), the equation becomes: \( 36960 = 40 \times 4\pi (R^2 - 49) \).
 
Simplifying, we get: \( 36960 = 160\pi (R^2 - 49) \).
 
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 calculated new radius of the soap bubble is \( 7 \, \text{cm} \).