A rectangular film of liquid is extended from $(4 ext{ cm} imes 2 ext{ cm})$ to $(5 ext{ cm} imes 4

Question

A rectangular film of liquid is extended from $(4 	ext{ cm} 	imes 2 	ext{ cm})$ to $(5 	ext{ cm} 	imes 4 	ext{ cm})$. If the work done is $3 	imes 10^{-4} 	ext{ J}$, then the value of the surface tension of the liquid is:

Accepted Answer

1. **Surface Energy Relation:** The work done ($W$) 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 ($T$) and the total change in surface area ($ \Delta A $). The formula is $ W = T 	imes \Delta A $ .
2. **Area Calculation:**
   - Initial Area, $ A_1 = 4 	ext{ cm} 	imes 2 	ext{ cm} = 8 	ext{ cm}^2 = 8 	imes 10^{-4} 	ext{ m}^2 $.
   - Final Area, $ A_2 = 5 	ext{ cm} 	imes 4 	ext{ cm} = 20 	ext{ cm}^2 = 20 	imes 10^{-4} 	ext{ m}^2 $.
   - Change in area (one side) = $ 20 - 8 = 12 	imes 10^{-4} 	ext{ m}^2 $.
3. **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:
   $$ \Delta A_{total} = 2 	imes (A_2 - A_1) = 2 	imes 12 	imes 10^{-4} = 24 	imes 10^{-4} 	ext{ m}^2 $$
4. **Calculation:**
   Substituting the values into the work formula:
   $$ 3 	imes 10^{-4} = T 	imes (24 	imes 10^{-4}) $$
   $$ T = \frac{3 	imes 10^{-4}}{24 	imes 10^{-4}} = \frac{3}{24} = \frac{1}{8} = 0.125 	ext{ Nm}^{-1} $$

Step-by-Step Solution

Practice All PHYSICS Questions