Question

A circular film of a liquid has an area of 10 cm$^2$. If the work done in making its radius two times the initial radius is $8 \times 10^{-3$ J, the surface tension of the liquid is $\left(1+\frac{1}{a}\right)$ Nm$^{-1}$. The value of $a$ is}

• A. 5
• B. 4
• C. 3
• D. 2

Hint:

A liquid film (e.g., soap film) has two surfaces. The total change in surface area is twice the change in the film's geometric area.
Work done against surface tension: $W = T \times \Delta A_{total}$.
Ensure area units are converted to m$^2$ for consistency with Joules and Nm$^{-1}$.

Answer 1

The Correct Option is B

A circular film of liquid (like a soap film on a loop) has two surfaces (top and bottom). Work done $W$ in changing the surface area by $\Delta A_{total}$ is $W = T_{s} \times \Delta A_{total}$, where $T_{s}$ is the surface tension. (Using $T_s$ to avoid confusion with Temperature). $\Delta A_{total} = 2 \times \Delta A_{film\_one\_surface}$ because there are two surfaces. Initial area of the film (one surface) $A_1 = 10 \text{ cm}^2 = 10 \times (10^{-2} \text{ m})^2 = 10 \times 10^{-4} \text{ m}^2 = 10^{-3} \text{ m}^2$. Let $r_1$ be the initial radius, so $A_1 = \pi r_1^2$. The radius is made two times the initial radius, so $r_2 = 2r_1$. The new area of the film (one surface) $A_2 = \pi r_2^2 = \pi (2r_1)^2 = 4 \pi r_1^2 = 4A_1$. $A_2 = 4 \times (10^{-3} \text{ m}^2) = 4 \times 10^{-3} \text{ m}^2$. Change in area for one surface: $\Delta A_{film\_one\_surface} = A_2 - A_1 = 4A_1 - A_1 = 3A_1$. $\Delta A_{film\_one\_surface} = 3 \times (10^{-3} \text{ m}^2) = 3 \times 10^{-3} \text{ m}^2$. Total change in surface area (considering two surfaces): $\Delta A_{total} = 2 \times \Delta A_{film\_one\_surface} = 2 \times (3 \times 10^{-3} \text{ m}^2) = 6 \times 10^{-3} \text{ m}^2$. Work done $W = 8 \times 10^{-3} \text{ J}$. So, $T_s = \frac{W}{\Delta A_{total}} = \frac{8 \times 10^{-3} \text{ J}}{6 \times 10^{-3} \text{ m}^2} = \frac{8}{6} \text{ Nm}^{-1} = \frac{4}{3} \text{ Nm}^{-1}$. We are given that the surface tension $T_s = \left(1+\frac{1}{a}\right) \text{ Nm}^{-1}$. Therefore, $\frac{4}{3} = 1 + \frac{1}{a}$. $\frac{1}{a} = \frac{4}{3} - 1 = \frac{4}{3} - \frac{3}{3} = \frac{1}{3}$. So, $a = 3$. This corresponds to option (c). However, the image indicates option (b) is correct, meaning $a=4$. If $a=4$, then $T_s = 1 + 1/4 = 5/4$. Then $W = T_s \times \Delta A_{total} = (5/4) \times (6 \times 10^{-3}) = (30/4) \times 10^{-3} = 7.5 \times 10^{-3}$ J. This does not match the given work done of $8 \times 10^{-3}$ J. The calculation for $a=3$ is consistent with the problem statement. There is a discrepancy with the marked correct answer. Assuming the calculation is correct: \[ \boxed{a=3} \] (Note: Solution adheres to calculation, marked answer implies $a=4$ which is inconsistent.)