AIIMS 2007 Physics Work and Energy MCQ Question

To determine the surface tension of the soap film based on the work done to increase its size, we first need to understand the relationship between work, surface tension, and the change in area of the soap film.

Relevant Concepts

Surface Tension ($\gamma$) is defined as the force per unit length acting at the surface of a liquid. For a soap film, which has two surfaces (front and back), the total surface area affects the amount of work done when the film's area is increased.

The work done ($W$) in changing the area of the film is given by the formula:

$$W = \gamma \Delta A$$

Where:

• $W$ is the work done,
• $\gamma$ is the surface tension,
• $\Delta A$ is the change in area of the film.

Step-by-Step Calculation

1. Calculate the Initial and Final Areas:

   • Initial dimensions: $10 \, \text{cm} \times 6 \, \text{cm} = 10 \times 6 = 60 \, \text{cm}^2$
   • Final dimensions: $10 \, \text{cm} \times 11 \, \text{cm} = 10 \times 11 = 110 \, \text{cm}^2$

2. Calculate the Change in Area ($\Delta A$):

   $$\Delta A = A_{\text{final}} - A_{\text{initial}} = 110 \, \text{cm}^2 - 60 \, \text{cm}^2 = 50 \, \text{cm}^2$$

   Convert this area from cm² to m²:

   $$\Delta A = 50 \, \text{cm}^2 \times \left(\frac{1 \, \text{m}}{100 \, \text{cm}}\right)^2 = 50 \times 10^{-4} \, \text{m}^2 = 5.0 \times 10^{-3} \, \text{m}^2$$

3. Use the Work Done to Find Surface Tension ($\gamma$): Given that the work done is $W = 3.0 \times 10^{-4} \, \text{J}$:

   Rearranging the formula for surface tension gives us:

   $$\gamma = \frac{W}{\Delta A}$$

   Plugging in the values:

   $$\gamma = \frac{3.0 \times 10^{-4} \, \text{J}}{5.0 \times 10^{-3} \, \text{m}^2}$$ $$\gamma = \frac{3.0}{5.0} \times 10^{-1} \, \text{N/m} = 0.6 \times 10^{-1} \, \text{N/m} = 6.0 \times 10^{-2} \, \text{N/m}$$

   However, since we only account for the two surfaces of the soap film:

   $$\gamma_{\text{effective}} = \frac{6.0 \times 10^{-2}}{2} = 3.0 \times 10^{-2} \, \text{N/m}$$

Thus, the surface tension of the soap film is:

$$\gamma = 3.0 \times 10^{-2} \, \text{N/m}$$

Conclusion

Correct Answer: B) $3 \times 10^{-2} \, \text{N/m}$

Why Other Options Are Incorrect

• A) $5 \times 10^{-2} \, \text{N/m}$: This value may have been derived from an incorrect application of the work-energy principle or calculating the effective area.
• C) $1.5 \times 10^{-2} \, \text{N/m}$: This value would suggest that the work done was significantly less effective than calculated.
• D) $1.2 \times 10^{-2} \, \text{N/m}$: Like option C, this value is not aligned with the relationship between work done and area change based on our calculations.

In conclusion, by using the correct calculations based on the given work and area changes, we establish that the surface tension of the soap film is indeed $3 \times 10^{-2} \, \text{N/m}$.