GATE (TF) Textile 2020 Question Paper Solution | GATE/2020/TF/55

Question 55 (Textile Engineering & Fibre Science)

The work of adhesion (WSL) depends on the surface tension (\gamma_{LV}) of the liquid and the contact angle (\theta) formed on a surface and is expressed as \gamma _{LV}(1+cos\theta)). The WSL for a given fabric and a liquid is reduced to 1/3rd of the original value after oil repellent treatment. If the measured contact angle of the untreated fabric is 60o, the percentage change in the contact angle after the treatment is __100__.

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Given in the question:

Work of adhesion(WSL)=\gamma _{LV}(1+cos\theta))

Where, (\gamma_{LV})=Surface tension of the liquid
WSL=Work of adhesion
and
Let ,WSL1=1 then, WSL2=(1/3)*WSL1 (After oil treatment)

Let,measured contact angle of untreated fabric is (\theta_1) & Measured contact angle of treated fabric is (\theta_2)
So ,(\theta_1)=60o

Percentage change in the contact angle after treatment(\Delta\theta)=?

Calculation:

Work of adhesion(WSL)=\gamma _{LV}(1+cos\theta))

i.e \frac{WSL_1}{WSL_2}=\frac{(1+cos\theta_1)}{(1+cos\theta_2)}

\frac{WSL_1}{(1/3)*WSL_1}=\frac{(1+cos60^o)}{(1+cos\theta_2)}

3=\frac{(1+1/2)}{(1+cos\theta_2)}

3=\frac{(3/2)}{(1+cos\theta_2)}

{(1+cos\theta _2)}= 3 \div (3/2)

{(1+cos\theta_2)}=3 \times \frac{2}{3}

{(1+cos\theta_2)}=2

{cos\theta_2}=2-1

cos\theta_2=1

\theta_2=Cos^{-1}1

\theta_2=0^o

Now, \Delta \theta = \frac{\theta_1 - \theta_2 }{\theta_1} \times 100


\Delta \theta = \frac{60-0}{60} \times 100

\Delta \theta = \frac{60}{60} \times 100

\Delta\theta=100 (Answer)









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