Question 228: Properties of Matter

Question

The surface tension of a soap solution is $3.5 	imes 10^{-2} \mathrm{~N} / \mathrm{m}$. The work required to increase the radius of a soap bubble from 1 cm to 2 cm is $\alpha 	imes 10^{-6} \mathrm{~J}$. The value of $\alpha$ is $\_\_\_\_$ .
$$ (\pi=22 / 7) $$

Accepted Answer

For a soap bubble, there are
two surfaces
, so the increase in surface energy is
$$ W = 2T \Delta A $$
where $T$ is the surface tension and $\Delta A$ is the increase in area of the bubble.
The surface area of a sphere is
$$ A = 4\pi r^2 $$
So, increase in area for one surface:
$$ \Delta A = 4\pi \left(r_2^2 - r_1^2ight) $$
Since a soap bubble has two surfaces, total work done is
$$ W = 2T \cdot 4\pi \left(r_2^2 - r_1^2ight) = 8\pi T \left(r_2^2 - r_1^2ight) $$
Now,
$$ r_1 = 1 	ext{ cm} = 10^{-2}	ext{ m} $$
$$ r_2 = 2 	ext{ cm} = 2	imes 10^{-2}	ext{ m} $$
So,
$$ r_2^2 - r_1^2 = (2	imes 10^{-2})^2 - (10^{-2})^2 $$
$$ = 4	imes 10^{-4} - 1	imes 10^{-4} = 3	imes 10^{-4} $$
Given,
$$ T = 3.5	imes 10^{-2}\ 	ext{N/m} $$
Thus,
$$ W = 8\pi 	imes 3.5	imes 10^{-2} 	imes 3	imes 10^{-4} $$
Using $\pi = \dfrac{22}{7}$,
$$ W = 8 	imes \frac{22}{7} 	imes 3.5 	imes 3 	imes 10^{-6} $$
Since $3.5 = \dfrac{7}{2}$,
$$ W = 8 	imes \frac{22}{7} 	imes \frac{7}{2} 	imes 3 	imes 10^{-6} $$
$$ = 8 	imes 11 	imes 3 	imes 10^{-6} $$
$$ = 264 	imes 10^{-6}\ 	ext{J} $$
Hence,
$$ \alpha = 264 $$
So, the answer is:
$$ \boxed{264} $$

Properties of Matter

The surface tension of a soap solution is 3.5 \times 10^{-2} \mathrm{~N} / \mathrm{m}. The work required to increase the radius of a soap bubble from 1 cm to 2 cm is \alpha \times 10^{-6} \mathrm{~J}. The value of \alpha is \_\_\_\_ . $ (\pi=22 / 7)

Integer / Numeric Answer Type