Q135 / 160 · AP EAMCET 2022 (07 Jul Shift 2) · Physics
A circular film of a liquid has an area of $10 \mathrm{~cm}^2$. If the work done in making its radius two times the initial radius is $8 \times 10^{-3} \mathrm{~J}$. the surface tension of the liquid is $\left(1+\frac{1}{\alpha}\right) \mathrm{Nm}^{-1}$. The value of $\alpha$ is
- A$5$
- B$4$
- Correct answerC$3$
- D$2$
Solution
If radius ( $r$ ) of a circle made two times (2r)
$$
\begin{aligned}
\text { then change in area } & =\pi(2 r)^2-\pi r^2 \\
& =3 \pi r^2=3 \times \text { initial area }
\end{aligned}
$$
Here, initial area of soap film, $A_1=10 \mathrm{~cm}^2$
Final area has a radius twice that of initial radius.
So, change in area
$$
\begin{aligned}
\Delta A & =3 A_1=3 \times 10 \mathrm{~cm}^2 \\
& =3 \times 10 \times 10^{-4} \mathrm{~m}^2 \\
& =3 \times 10^{-3} \mathrm{~m}^2
\end{aligned}
$$
As, a liquid film has two surfaces (one top and other bottom).
Change in surface energy or work done,
$$
\begin{aligned}
W & =2 \Delta A \times T \\
T & =\text { Surface tension } \\
T & =W / 2 \Delta A
\end{aligned}
$$
Here, $W=8 \times 10^{-3} \mathrm{~J}$
$$
\Delta A=30 \times 10^{-3} \mathrm{~m}^2
$$
So, surface tension of Liquid, $T=\frac{8 \times 10^{-3}}{2 \times 3 \times 10^{-3}}$
$$
=\frac{8}{6}=\frac{4}{3}=1+\frac{1}{3} \mathrm{Nm}^{-1}
$$
Hence, $\alpha=3$
Hence, the correct answer is option: 'C'.