First slide

Surface tension

Question

A rectangular film of liquid is extended from $(4 c m \times 2 c m)$ to $(5 c m \times 4 c m)$ . If the work done is $3 \times 10^{- 4} J$ the value of the surface tension of the liquid is

Easy

A

$0.250 N m^{- 1}$
B

$0.125 N m^{- 1}$
C

$0.2 N m^{- 1}$
D

$8 N m^{- 1}$

Solution

Work done = Surface tension of film $\times$ Change in area of the film
$or, W = T \times Δ A$
$Here, A_{1} = 4 cm \times 2 cm = 8 {cm}^{2}$
$A_{2} = 5 cm \times 4 cm = 20 {cm}^{2}$
$Δ A = 2 (A_{2} - A_{1}) = 24 {cm}^{2} = 24 \times 10^{- 4} m^{2}$
$W = 3 \times 10^{- 4} J, T = ?$
$∴ T = \frac{W}{Δ A} = \frac{3 \times 10^{- 4}}{24 \times 10^{- 4}} = \frac{1}{8} = 0.125 N m^{- 1}$

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