A soap bubble is blown to a diameter of 7 cm. if 36960 ergs of work is done in blowing if further find the new radius, if surface tension of the soap solution is 40 dynes/cm.

Explanation

To solve the problem, we will use the relationship between work done in blowing a soap bubble, surface tension, and the change in surface area.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Initial diameter of the soap bubble, \( d_1 = 7 \, \text{cm} \)
- Initial radius, \( r_1 = \frac{d_1}{2} = \frac{7}{2} = 3.5 \, \text{cm} \)
- Work done, \( W = 36960 \, \text{ergs} \)
- Surface tension, \( \gamma = 40 \, \text{dynes/cm} \)

2. **Calculate the Initial Surface Area:**
The surface area \( A_1 \) of a soap bubble (which has two surfaces) is given by:
\[ A_1 = 2 \times 4\pi r_1^2 = 8\pi r_1^2 \]
Substituting \( r_1 = 3.5 \, \text{cm} \):
\[ A_1 = 8\pi (3.5)^2 = 8\pi (12.25) = 98\pi \, \text{cm}^2 \]

3. **Calculate the Work Done in Terms of Surface Tension and Change in Surface Area:**
The work done \( W \) in blowing the bubble can be expressed as:
\[ W = \gamma \times \Delta A \]
where \( \Delta A = A_2 - A_1 \) and \( A_2 \) is the final surface area of the bubble with radius \( r_2 \):
\[ A_2 = 2 \times 4\pi r_2^2 = 8\pi r_2^2 \]
Therefore,
\[ \Delta A = 8\pi r_2^2 - 98\pi \]

4. **Set Up the Equation for Work Done:**
Substituting into the work done equation:
\[ 36960 = 40 \times (8\pi r_2^2 - 98\pi) \]
Simplifying:
\[ 36960 = 320\pi r_2^2 - 3920\pi \]
Rearranging gives:
\[ 320\pi r_2^2 = 36960 + 3920\pi \]

5. **Calculate \( r_2^2 \):**
Dividing both sides by \( 320\pi \):
\[ r_2^2 = \frac{36960 + 3920\pi}{320\pi} \]

6. **Calculate \( r_2 \):**
To find \( r_2 \), we need to compute the right-hand side. First, calculate \( 3920\pi \) (using \( \pi \approx 3.14 \)):
\[ 3920\pi \approx 12300.8 \]
Thus,
\[ 36960 + 12300.8 \approx 49260.8 \]
Now substituting back:
\[ r_2^2 \approx \frac{49260.8}{320 \times 3.14} \approx \frac{49260.8}{1005.6} \approx 48.93 \]
Therefore,
\[ r_2 \approx \sqrt{48.93} \approx 7 \, \text{cm} \]

### Final Result:
The new radius \( r_2 \) of the soap bubble is approximately \( 7 \, \text{cm} \).