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

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 $7 \, \text{cm}$.