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: \( \left( \pi = \frac{22}{7} \right) \).
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 \( 7 \, \text{cm} \).
Was this answer helpful?
0
1
Hide Solution
Verified By Collegedunia
Approach Solution -2
Step 1: Given data.
Initial diameter of soap bubble = \( 7 \, \text{cm} \Rightarrow r_1 = \frac{7}{2} = 3.5 \, \text{cm} \)
Work done in blowing = \( W = 36960 \, \text{erg} \)
Surface tension = \( T = 40 \, \text{dyne/cm} \)
We have to find the new radius \( r_2 \).
Step 2: Concept used.
For a soap bubble, both the inner and outer surfaces contribute to the surface energy.
Hence, work done in increasing the size of a soap bubble is:
\[
W = 8\pi T (r_2^2 - r_1^2)
\]
where \( r_1 \) is the initial radius and \( r_2 \) is the final radius.
