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NEET PHYSICSEasy

A rectangular film of liquid is extended from (4 cm×2 cm)(4 \text{ cm} \times 2 \text{ cm}) to (5 cm×4 cm)(5 \text{ cm} \times 4 \text{ cm}). If the work done is 3×10−4 J3 \times 10^{-4} \text{ J}, then the value of the surface tension of the liquid is:

A

0.250 Nm−10.250 \text{ Nm}^{-1}

B

0.125 Nm−10.125 \text{ Nm}^{-1}

C

0.2 Nm−10.2 \text{ Nm}^{-1}

D

8.0 Nm−18.0 \text{ Nm}^{-1}

Step-by-Step Solution

  1. Surface Energy Relation: The work done (WW) to increase the surface area of a liquid film is equal to the increase in surface energy, which is given by the product of surface tension (TT) and the total change in surface area (ΔA\Delta A). The formula is W=T×ΔAW = T \times \Delta A .
  2. Area Calculation:
  • Initial Area, A1=4 cm×2 cm=8 cm2=8×10−4 m2A_1 = 4 \text{ cm} \times 2 \text{ cm} = 8 \text{ cm}^2 = 8 \times 10^{-4} \text{ m}^2.
  • Final Area, A2=5 cm×4 cm=20 cm2=20×10−4 m2A_2 = 5 \text{ cm} \times 4 \text{ cm} = 20 \text{ cm}^2 = 20 \times 10^{-4} \text{ m}^2.
  • Change in area (one side) = 20−8=12×10−4 m220 - 8 = 12 \times 10^{-4} \text{ m}^2.
  1. Accounting for Two Surfaces: A liquid film (like a soap film) has two free surfaces (top and bottom). Therefore, the total increase in effective surface area is twice the geometric increase: ΔAtotal=2×(A2−A1)=2×12×10−4=24×10−4 m2\Delta A_{total} = 2 \times (A_2 - A_1) = 2 \times 12 \times 10^{-4} = 24 \times 10^{-4} \text{ m}^2
  2. Calculation: Substituting the values into the work formula: 3×10−4=T×(24×10−4)3 \times 10^{-4} = T \times (24 \times 10^{-4}) T=3×10−424×10−4=324=18=0.125 Nm−1T = \frac{3 \times 10^{-4}}{24 \times 10^{-4}} = \frac{3}{24} = \frac{1}{8} = 0.125 \text{ Nm}^{-1}
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