Radius of a soap bubble is changed from 7 cm to 14 cm then the work done in this process is (15000 − x) J. Find the value of x.

Given:

π = 22/7

σ = 0.04 N/m

The correct option is 216.

Step-by-step Explanation:

A soap bubble has two free surfaces in contact with air (inner and outer surfaces). Therefore, the work done in changing the radius of the soap bubble is given by the formula:
$W=2\times \sigma \times \Delta A$
where:
• $\sigma$ is the surface tension of the soap solution.
• $\Delta A$ is the change in the surface area of one side of the bubble.

The initial surface area of one side of the bubble is:
$A_1=4\pi R_1^2$
The final surface area of one side of the bubble is:
$A_2=4\pi R_2^2$
Thus, the total work done is:
$W=2\times \sigma \times 4\pi (R_2^2-R_1^2)=8\pi \sigma (R_2^2-R_1^2)$

Given data:
• Initial radius, $R_1=7\text{cm}=0.07\text{m}$
• Final radius, $R_2=14\text{cm}=0.14\text{m}$
• Surface tension, $\sigma =0.04\text{N/m}$
• $\pi =\frac{22}{7}$

Substitute these values into the equation:
$W=8\times \frac{22}{7}\times 0.04\times \left[\right(0.14{)}^2-\left(0.07{)}^2\right]$

Using scientific notation for calculations:
$W=8\times \frac{22}{7}\times 0.04\times {10}^{-4}\times ({14}^2-7^2)$
$W=\frac{176}{7}\times 0.04\times {10}^{-4}\times (196-49)$
$W=\frac{176}{7}\times 0.04\times {10}^{-4}\times 147$
Since $\frac{147}{7}=21$:
$W=176\times 0.04\times 21\times {10}^{-4}$
$W=176\times 0.84\times {10}^{-4}$
$W=147.84\times {10}^{-4}\text{J}$
$W=14784\times {10}^{-6}\text{J}$

The work done is expressed as $(15000-x)\times {10}^{-6}\text{J}$:
$15000-x=14784$
$x=15000-14784$
$x=216$