A soap bubble is blown to a diameter of 7 cm. 36960 erg of work is done in blowing it further. If surface tension of soap solution is 40 dyne/cm then the new radius is ______ cm. Take : ( π = 22/7)

The correct answer is 7.

Step-by-step Explanation:

A soap bubble has two free liquid-gas interfaces (an inner surface and an outer surface). Therefore, any change in the surface area of a soap bubble requires twice the work compared to a single-surface liquid drop.

The relation between the work done ($W$), surface tension ($T$), and change in surface area ($\Delta A$) is given by:
$W=2\times T\times \Delta A$
where $\Delta A$ is the increase in the outer surface area of the bubble.

Let $r_1$ be the initial radius and $r_2$ be the new radius of the bubble.
Given:
Initial diameter, $d_1=7\text{cm}$
Initial radius, $r_1=\frac{7}{2}\text{cm}=3.5\text{cm}$
Work done, $W=36960\text{erg}$
Surface tension, $T=40\text{dyne/cm}$
Value of $\pi =\frac{22}{7}$

The change in surface area is:
$\Delta A=4\pi r_2^2-4\pi r_1^2=4\pi (r_2^2-r_1^2)$

Substituting this back into the work equation:
$W=8\pi T(r_2^2-r_1^2)$

Now, let's substitute the given values to find $r_2$:
$36960=8\times \frac{22}{7}\times 40\times (r_2^2-{3.5}^2)$
$36960=\frac{7040}{7}\times (r_2^2-12.25)$

Rearranging the equation to solve for $(r_2^2-12.25)$:
$r_2^2-12.25=\frac{36960\times 7}{7040}$
$r_2^2-12.25=5.25\times 7$
$r_2^2-12.25=36.75$

Solving for $r_2^2$:
$r_2^2=36.75+12.25$
$r_2^2=49$
$r_2=\sqrt{49}=7\text{cm}$

Thus, the new radius of the soap bubble is 7 cm.