PYQ NEET- Mechanical Properties Of Fluids L-4

Question: A rectangular film of liquid is extended from $(4 \mathrm{~cm} \times 2 \mathrm{~cm})$ to $(5 \mathrm{~cm} \times 4 \times \mathrm{cm})$. If the work done is $3 \times 10^{-4} \mathrm{~J}$, the value of the surface tension of the liquid is#

A) $0.250 \mathrm{Nm}^{-1}$

B) $0.125 \mathrm{Nm}^{-1}$

C) $0.2 \mathrm{Nm}^{-1}$

D) $8.0 \mathrm{Nm}^{-1}$

Answer: $0.125 \mathrm{Nm}^{-1}$#

Sol:#

Key Idea Increase in surface energy= work done in area $\times$ surface tension $\because$ Increase in surface area, $$ \begin{aligned} & \Delta A=(5 \times 4-4 \times 2) \times 2 \ & \quad(\because \text { film has two surfaces }) \ & =(20-8) \times 2 \mathrm{~cm}^2=24 \mathrm{~cm}^2 \ & =24 \times 10^{-4} \mathrm{~m}^2 \end{aligned} $$

So, work done, $W=T \cdot \Delta \mathrm{A}$ $$ \begin{aligned} & 3 \times 10^{-4}=T \times 24 \times 10^{-4} \ \therefore \quad & T=\frac{1}{8}=0.125 \mathrm{~N} / \mathrm{m} \end{aligned} $$