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 \( 7 \, \text{cm} \).
Two soap bubbles of radius 2 cm and 4 cm, respectively, are in contact with each other. The radius of curvature of the common surface, in cm, is _______________.

Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).
