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Physics Question on mechanical properties of fluid

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

A

0.250 Nm−10.250 \, N m^{-1}

B

0.125 Nm−10.125 \, N m^{-1}

C

0.2 Nm−10.2 \, N m^{-1}

D

8.0 Nm−18.0\, N m^{-1}

Answer

0.125 Nm−10.125 \, N m^{-1}

Explanation

Solution

W=2(TΔA)W = 2\left(T\Delta A\right)
ΔA=A2−A1\Delta A = A_{2} -A_{1}
=(20−8)×10−4m2= \left(20 - 8 \right) \times10^{-4}m^{2}
=12×10−4m2= 12 \times10^{-4} m^{2}
T=W2ΔA=3×10−42×12×10−4T = \frac{W}{2\Delta A} = \frac{3\times10^{-4}}{2 \times12\times10^{-4}}
T=18Nm=0.125NmT = \frac{1}{8} \frac{N}{m} = 0.125 \frac{N}{m}